Title: Part IB / Linear Algebra
Tags: Bilinear_Forms Linear_Maps Vector_Spaces Endomorphisms Inner_Product_Spaces Determinants Dual_Spaces Adjugates

Q: Definition.  Let F be an arbitrary field. What is an F-vector space?
A: An abelian group \((V, +)\) equipped with a function \[F \times V \to V \; \; (\lambda, v) \mapsto \lambda v\]
which satisfies
i) \(\lambda(v_1 + v_2) = \lambda v_1 + \lambda v_2\)
ii) \((\lambda_1 + \lambda_2)v = \lambda_1 v + \lambda_2 v\)
iii) \(\lambda(\mu v) = (\lambda \mu) v\)
iv) \(1v = v\)
Such a vector space may also be called a vector space over F.

Q: Definition. Linear Dependence. We say that \(v_1, \dots , v_n \in V\) are linearly independent if, for \(\lambda_i \in F\) [...]
A: \[\sum_{i=1}^n \lambda_i v_i = 0 \implies \forall i, \lambda_i = 0\]
Back Extra: Linear dependence if there are finitely many vectors we can make a linear combination of and form zero.
If \(0 \in S\), then S is linearly dependent as \(1 \cdot 0 = 0\)

Q: Definition. Span. Let \(S \subseteq V\), the span of S is [...]
A: the set of all linear combination of elects in S. We denote this as <S> \[ <S> = \left \{ \sum_{i=1}^k \alpha_i v_i : k = 1,2,\dots;  \alpha_1, \dots, \alpha_k \in F; v_1, \dots, v_k \in S \right \} \]
Back Extra: 0 in span for all non-empty S.

Q: Definition. Basis. Let V be a vector space. \(B \subseteq V\) is a basis for C if
[...]
A: i) B is linearly independent
ii) <B> = V

Q: Proposition. V a vector space. \(B \subseteq V\) is a basis for V iff every vector in V can be uniquely written as a linear combination of elements of B.
A: (->) Let \(v \in V, v = \sum_{i=1}^k \alpha_i v_i\) and \(v = \sum_{i=1}^k \beta_i v_i\). So subtract one from other to get 0, since B is a basis this is a contradiction unless all coefficients equal. Since B spans we have unique linear combination for each v.
(<-) We know B spans V. If B wasn't linearly independent, the linear combinations would not be unique, for example we could make 0 in two ways. Therefore B must be basis.

Q: Lemma. Steinitz Exchange Lemma. Let V be a finite dimensional F-vector space. Let \((v_1, \dots , v_m)\) be linearly independent, and \((w_1, \dots , w_n)\) which spans V. Then,
i) \(m \leq n\)
ii) up to reordering, \((v_1, \dots, v_m, w_{m+1}, \dots , w_n)\) spans V.
A: Suppose we have replaced \(l \geq 0\) of the \(w_i\). \[< v_1, \dots, v_l, w_{l+1}, \dots w_n> = V\]
If m=l, we are done. Otherwise \(l < m\). Then, \(v_{l+1} \in V = < v_1, \dots, v_l, w_{l+1} \dots, w_n>\) Hence we can express \(v_{l+1}\) as a linear combination of the generating set. Since the \((v_i)_{1 \leq i \leq m}\) are linearly independent, one of the coefficients on the \(w_i\) are nonzero. In particular, up to reordering we can express \(w_{l+1}\) as a linear combination of \(v_1, \dots , v_{l+1}, w_{l+2}, \dots, w_n\). Inductively we may replace m of the w terms with v terms, Since we have replaces m vectors, necessarily \(m \leq n\).

Q: Lemma. Let V be a vector space.
i) If \(v_1, \dots , v_n\) span V then there exists a subset of them that is a basis for V.
ii)If \(v_1, \dots, v_n\) is L.I. and S spans V then \(\exists v_{n+1}, \dots , v_N \in S\) s.t. \(v_1, \dots, v_n , v_{n+1}, \dots, v_N\) is a basis for V.
A: i) Consider a linearly independent subset of \(\{v_i\}_{i=1}^n\) and w.l.o.g assume \(v_1, \dots , v_m\) is such a set. For all \(i = m+1, \dots, n\) we have \[\sum_{k=1}^m \lambda_j^i v_j + \lambda_i v_i = 0\] since \(v_1, \dots, v_m, v_i\) is linearly dependent. (hence \(\lambda_1^i, \dots, \lambda_m^i, \lambda_i\) are not all zero).
Also \(\lambda_i \neq 0\)  in such case otherwise \(v_1,\dots,v_m\) would be linearly dependent. So \[v_i = \frac{-1}{\lambda_i}\sum_{j=1}^m \lambda_j^i v_i\]
So \(V = <\{v_1, \dots, v_n\}> \subset <\{ v_1, \dots, v_m\}>\) and we are done.
ii) First apply Steinitz Exchange Lemma to find \(v_1, \dots v_n, v_{n+1} \dots v_{N'}\) that spans V.
Now choose a maximal L.I. subset which contains \(v_1, \dots , v_n\). This is possible because those vectors are linearly independent. Then proceeds as in (i)

Q: Definition. Finite Dimensional. A vector space V is finite dimensional if [...]
A: there exists a finite spanning set.

Q: Lemma. Let V be finite dimensional.
i) V has a finite basis.
ii) all bases of V have the same size.
A: i)Apply lemma to our finite spanning set to form a basis.
ii) If \(\mathcal{B}= \{x_i\}_i\) is an infinite basis. Simply consider \(x_1, \dots , x_n\) where \(v_1, \dots ,v_n\) is a finite basis of V. Now apply Steinitz Exchange Lemma to \(x_1 \dots x_n\)(L.I) and \(v_1 \dots v_n\)(which spans) to see \(x_1, \dots ,x_n\) spans and thus is a basis.
Now let \(v_1, \dots v_n\) and \(w_1, \dots w_m\) be bases which are linearly independent and span respectively so \(n \leq m\) by Steinitz.
Applying the same thing the other way gives \(m \leq n \implies n = m\).

Q: Definition. Dimension. If V is a vector space, finite dimensional, define [...]
A: \[\dim V = \text{ size of a basis}\]
Back Extra: This is well defined as all bases will have the same size.

Q: Lemma. V a vector space. \(\dim V = n\)
i) \(v_1, \dots ,v_n\) linearly independent \(\implies v_1 \dots v_n\) are a basis
ii)If \(S \subset V\) and S spans V then \(|S| \geq n\) and equality holds iff S is a basis.
A: i) so there exists \(x_1 \dots x_n\) that span V. Steinitz Exchange Lemma with \(v_1 \dots v_n\) and \(x_1 \dots x_n\) to see that \(v_1 \dots v_n\) span.
ii) Let \(\overline{S} \subseteq S\) be a basis of V. By definition \(|\overline{S}| = n\) so we are done.
By definition \(S\) basis \(\implies |S| = n\)
\(|S| = n\) if \(\overline{S} \subset S\) is a basis then we have a basis of size less than n. Contradiction.

Q: Definition. Subspace. We say that \(W \subseteq V\), when V is a vector space, is a subspace if
i) [...]
ii) [...]
iii)[...]
A: closed under addition; closed under scalar multiplication; non-empty
Back Extra: If U is a subspace of V then we write \(U \leq V\)

Q: Lemma. Let \(U_1, \dots , U_n \leq V\), V a vector space then \(W = \cap_{i=1}^n U_i\) is a subspace.
A: Just show axioms.

Q: Definition. Let \(U_1 , \dots U_n \leq V\), V a vector space, then \[\sum_{i=1}^k U_i = [...]\]
A: \{ u_1 + \dots + u_k : u_1 \in U_1, \dots , u_k \in U_k\}
Back Extra: The sum of subspace is a subspace.

Q: Theorem. Let \(U,W \leq V\), V is a vector space which is finite dimensional. then U+W and \(U \cap W\) are finite dimensional and \[\dim (U+W) = \dim U + \dim W - \dim (U \cap W)\]
A: Let \(v_1 \dots v_k\) be a basis for \(U \cap W\)
Let \(v_1 \dots v_k u_{k+1} \dots u_n\) be a basis for U
Let \(v_1 \dots v_k w_{k+1} \dots w_m\) be a basis for W
We want to show \(S = {v_1 \dots v_k u_{k+1} \dots u_n w_{k+1} \dots w_m}\) is a basis for \(U + W\).
Easy to show it spans.
For linear independence assume \(\alpha_i, \beta_i, \gamma_i\) such that \[\sum_{i=1}^k \alpha_i v_i + \sum_{i=k+1}^n \beta_i u_i + \sum_{i = k+1}^m \gamma_i w_i = 0\]
So \[x = \sum_{i = 1}^k \alpha_i v_i + \sum_{i = k+1}^n \beta_i u_i = - \sum_{i k+1}^m \gamma_i w_i\]
Middle is in U, RHS in W so x in the intersection and can write \(x = \sum \delta_i v_i = \sum (- \gamma_i)w_i\) but \(v_1 \dots v_k w_{k+1} \dots w_m\) are linearly independent so all coefficients 0. Do the same for v's and u's. So S is linearly independent and is a basis.

Q: Lemma. Let \(U \leq V\), V a vector space. If V is finite dimensional, then U is finite dimensional.
A: Let \(x_1 \dots x_n\) be a basis for V. Assume for contradiction that U is not finite dimensional. Then for each \(m \in \mathbb{N}\) I can construct \(u_1 \dots u_m \in U\) that are linearly independent which implies they are linearly independent in V which is a contradiction.

Q: Definition. Let \(U_1 , \dots, U_k \leq V\), V a vector space, we say \(\sum_{i=1}^k U_i\) is a direct sum [...] We will write, in this case, \(\bigoplus U_i\)
A: if \(\forall u \in U_1 + \dots + U_k\), u can be written uniquely as \(u = u_1 + \dots + u_k\) where \(u_i \in U_i\).

Q: Lemma. Let \(U, W \leq V\) then \(U+W\) is a direct sum iff \(U \cap W = \{0\}\)
A: If \(U + W\) is a direct sum and \(x \in U \cap W\), then \(x + 0 = x = 0 + x\) which is not unique unless \(x = 0\).
If \(U \cap W = \{0\}\)  and assume \(x = u+w = u' + w'\), where \(u,u' \in U, w,w' \in W\). Now we have \( u - u' = w - w'\) with LHS in U and RHS in W so we must have \(u = u'\) and \(w = w'\).

Q: Definition. Let \(U \leq V\). We say \(W \leq V\) is a complement of U if [...]
A: \(U \bigoplus W = V\).

Q: Proposition. Let V be a vector space, \(U \leq V\). If V is finite dimensional then
i) \(\dim U \leq \dim V\), '=' holds iff \(U = V\).
ii) U has a complement in V
A: i) U is finite dimensional. So let \(u_1, \dots , u_k\) be a basis for U. Now extend this basis to a basis of V : \(u_1 \dots u_k u_{k+1} \dots u_n\). Then \(\dim U \leq \dim V\), if \(k = n\) then \(U = <u_1 \dots u_k> = <u_1 \dots u_n> = V\).
ii) Let \(W = <u_{k+1} \dots u_n>\). We claim W is a complement of U in V. Note that \(U \cap W = \{0\}\). Let x be in the intersection, can write it in terms of u or w, so we can create a linear combination that gives 0, which is only possible if x = 0.

Q: Definition. Let V,W be vector spaces. We say that a function \(T : V \to W\) is linear map if [...]
A: a) \(T(u+v) = T(u) + T(v) \; \; \forall u,v \in V\)
b) \(T(\lambda u) = \lambda T(u) \;\; \forall u \in U, \lambda \in F\)

Q: Definition. We say that \(T : V \to W\), a linear map, is a isomorphism if [...] We say that T is an automorphism if [...]
A: T is also a bijection.; V = W

Q: Lemma. If \(T: V \to W\) is an isomorphism then \(T^{-1}: W \to V\) is an isomorphism.
A: It is enough to show that \(\forall \lambda_1 , \lambda_2 \in \mathbb{F}, y_1, y_2 \in W\) we have \[T^{-1}(\lambda_1 y_1 + \lambda_2 y_2) = \lambda_1 T^{-1}(y_1) + \lambda_2 T^{-1}(y_2)\]
Let \(x_1, x_2\) be such that \(T x_1 = y_1\) and \(T x_2 = y_2\). So \[\lambda_1 y_1 + \lambda_2 y_2 = T(\lambda_1 x_1 + \lambda_2 x_2)\]. So subbing into the thing from before gives the result.

Definition. Let \(T : V \to W\) be a linear map, we define \(\text{Ker}(T) =  \{v \in V : Tv = 0\} \). Define \(\text{null}(T) =  \dim (\text{Ker}(T))\).
Define \(\text{im}(T) =  \{Tv : v \in V\} \). Define \(\text{rank}(T) =  \dim (\text{im}(T))\)
Back Extra: Kernel is subspace of V, image is subspace of W.

Q: Theorem. Rank-Nullity Theorem. Let \(V,W\) be finite dimensional spaces. Let \(T : V \to W\) we a linear map, then \(\dim V = \text{rank} (T) + \text{null}(T)\)
A: Let \(x_1 \dots x_k\) be a basis for the kernel of T, now extend to a basis of V, \(x_1 \dots x_k x_{k+1} \dots x_n\). Now we can show \(T(x_{k+1}) \dots T(x_n)\) are a basis for \(\text{im}(T)\). Let \(y \in \text{im}(T)\) then \(\exists x\) s.t. \(T(x) = y\). Now write \(x = \sum_{i=1}^n \alpha_i x_i\), then \(y = \sum_{i = k+1}^n \alpha_i T(x_i)\). So \(T(x_{k+1}), \dots , T(x_n)\) spans \(\text{im} (T)\).
Assume \(\sum_{i = k+1}^n \alpha_i T(x_i) = 0\) so \(\sum_{i=k+1}^n \alpha_i x_i = \sum_{i=1}^k \beta_i x_i\) as it must be in the kernel. Since \(x_1 \dots x_n\) are L.I. we have all the coefficients being 0 so the \(T(x_{k+1}), \dots, T(x_n)\) are linearly independent and therefore a basis.

Q: Corollary. Let \(V,W\) be finite dimensional spaces over \(\mathbb{F}\). Let \(T : V \to W\) be a linear map, \(\dim V = \dim W\). Then T is an isomorphism iff T is surjective(injective).
A: The forward direction is trivial.
In the other direction we have \(\dim V = \text{rank}(T) + \text{im}(T)\) but T surjective so \(\text{null}(T) = 0 \implies Tx = Ty \implies x = y \implies T \text{ injective and therefore isomorphism} \).
Proof also works if we assume injectivivity.

Q: Lemma. Let \(V,W\) be vector spaces over \(\mathbb{F}\). Let \(\mathcal{B}\) be a basis for V and let \[T_0 : \mathcal{B} \to W\] then there exists a unique linear map \[T : V \to W\] s.t. \(Tx = T_0 x\) for all \(x \in \mathcal{B}\). "A linear map is determined uniquely by what it does to a basis".
A: Let \(x \in V\) and write \(x = \sum_{i=1}^n \alpha_i x_i\) where \(x_i \in \mathcal{B}, \alpha_i \in \mathbb{F}\). Then define \[T(x) = \sum_{i=1}^n \alpha_i T_0(x_i)\]. Need to check linearity, well-definedness and uniqueness.

Q: Theorem. Let \(V,W\) be finite dimensional vector spaces over \(\mathbb{F}\). Then \[V \cong W  \iff \dim V = \dim W\]
"Finite dimensional spaces are determined, up to isomorphism, by their dimension"
A: (->)If \(x_1 \dots x_n\) is a basis of V then \(Tx_1 \dots Tx_n\) is a basis of W, where \(T : V \to W\) is an isomorphism.
(<-)Let \(x_1 \dots x_n\) be a basis of V and let \(w_1 \dots w_n\) be a basis of W and define a function \(T_0 : \{x_0 , \dots , x_n\} \to W\) by \[T_0 x_i = w_i\]So extended to a linear map \(T : V \to W\) for which \(T x_i = w_i\). Clearly surjective and therefore an isomorphism.

Corollary. If V is finite dimensional over \( \mathbb{F}\) then \(V \cong \mathbb{F}^n\), where \(n = \dim V\).

Q: Definition. Let \(U \leq V\), where V is a vector space over \(\mathbb{F}\). We define the quotient space \(V/U\) to be [...].
A: the set of cosets of U in V. \(\{x + U : x \in V\}\). We define scalar multiplication by \[\lambda \cdot (x + U) = \lambda \cdot x + U\] for all \(\lambda \in \mathbb{F}\)

Q: Definition. Let \(T:V \to W\) be a linear map between finite dimensional vector spaces. Let \(\mathcal{B}\) be a basis for V and \(\mathcal{C}\) be a basis for W. We define the matrix \([T]_{\mathcal{B,C}}\) to be \((a_{i,j})_{i,j}\) where [...]
A: \[T v_j = \sum_{i=1}^m a_{ij}w_i\]
Back Extra: where \(\mathcal{B} = \{v_1 \dots v_n\}\) and \(\mathcal{C} = \{w_1 \dots w_n\}\)

Q: Lemma. Let U,V,W be finite dimensional vector spaces over \(\mathbb{F}\). Let \(\mathcal{A,B,C}\) be bases for U,V,W respectively and \[U \to^{T} V \to^{S} W\] where T,S are linear maps. Then \[[S \cdot T]_{\mathcal{A,C}} = [S]_{\mathcal{B,C}} \cdot [T]_{\mathcal{A,B}}\]
A: Let \(\mathcal{A} = \{ u_1 \dots u_p\}, \mathcal{B} = \{v_1 \dots v_n\} , \mathcal{C} = \{w_1 \dots w_m\}\). Let \(A= (a_{ij})_{ij}\) be \([T]_{\mathcal{A,B}}\) and \(B = (b_{ij})_{ij} = [S]_{\mathcal{B,C}}\). \[T u_j = \sum_{i=1}^n a_{ij}v_i \; \; \; \text{for each } 1 \leq j \leq p\] \[S v_i = \sum_{k=1}^m b_{ki}w_k \; \; \; \text{for each } 1 \leq i \leq m\] So \(S(T u_j) = S(\sum_{1=1}^n a_{ij}v_i) =. \sum_{i=1}^n a_{ij}S v_i = \sum_{i=1}^n a_{ij}(\sum_{k=1}^m b_{ki} w_k) = \sum_{k=1}^m \left [ \sum_{i=1}^n a_{ij}b_{ki} \right ] \cdot w_k = \sum_{k=1}^m \left ([S]_{\mathcal{B,C}} \cdot [T]_{\mathcal{A,B}} \right )_{kj}w_k\).

Q: Definition. Let \(V,W\) be vector spaces over \(\mathbb{F}\). Let \(L(V,W)\) be [...]
A: the set of linear maps \(T : V \to W\).
Back Extra: We endow \(L(V,W)\) with a vector space structure:
\(T+S\) is the linear map \(x \mapsto Tx + Sx\) for all x.
\(\lambda \cdot T\) is the linear map \(x \mapsto \lambda \cdot Tx\)

Q: Definition. We define \(\text{Mat}_{m,n} (\mathbb{F})\) to be [...]
A: the set of \(m \times n\) matrices over \(\mathbb{F}\).

Q: Proposition. Let \(V,W\) be vector spaces over \(\mathbb{F}\), \(\dim V = n, \dim W = m\). Let \(\mathcal{B}\) be a basis for V and \(\mathcal{C}\) be a basis for W then the map \[\theta (T) = [T]_{\mathcal{B, C}}\] for all \(T \in L(V,W)\) is an isomorphism. A corollary of this is \[\dim (L(V,W)) = (\dim V)(\dim W)\]
A: We need to check linear + bijective. Let \(\mathcal{B} = \{v_1 \dots v_n\}, \mathcal{C} = \{w_1 \dots w_m\}\). We need to show \[[T+S]_{\mathcal{B,C}} = [T]_{\mathcal{B,C}} + [S]_{\mathcal{B,C}}\]. If we let \((a_{ij})\) be the matrix for T and \((b_{ij})\) be the matrix for S. Then \[(T+S)v_i = T v_i + S v_i = \sum_{j=1}^m (a_{ji} + b_{ji})w_i\] so we have shown it. Easy to check also preserves scalar multiplication and therefore linear.
Injective:Since \(\theta\) is linear just need to check \(\theta(T) = 0 \implies T=0\) which means that every basis vector in \(\mathcal{B}\) is mapped to zero by T. The zero map also does this so \(T = 0 \in L(V,W)\).
Surjective: Consider the linear maps \(T_{i,j} \; 1 \leq i \leq n \; 1 \leq j \leq m\) \[T_{i,j}v_i = w_j \text{ and }T_{i,j} v_k = 0 \text{ for } k \neq i\]. \(\theta(T_{ij})\) is 1 only at \(i,j\) so \(\{T_{i,j}\}\) maps to the standard basis.

Q: Lemma. Let \(\alpha , \beta : V \to V\) be linear maps on a finite dimensional vector space. Then \[\alpha \circ \beta = \text{id}_V \iff \beta \circ \alpha = \text{id}_V\]and \(\alpha^{-1} = \beta\) if either holds.
A: \(\alpha \circ \beta = \text{id}_V \implies \alpha \text{ is surjective and therefore isomorphism} \implies \alpha^{-1} \text{ exists}\) and applying \(\alpha^{-1}\) to both sides gives the result, then using results about inverses gives us the first result.

Q: Lemma. Let \(A \in \text{Mat}_{m,n}(\mathbb{F})\). Then \[\exists B : AB = I_n \iff \exists C : CA = I_n\] if either holds then \(B = C\) is unique and we denote this matrix \(A^{-1}\) and say A is invertible.
A: (->) \(AB = I_n\) now for basis \(\mathcal{B}\)of \(\mathbb{F}^n\) and let \(\theta\) be the corresponding isomorphism \[\theta :(\mathbb{F}^n , \mathbb{F}^n) \to \text{Mat}_{n,n}(\mathbb{F})\] so defined by \(\alpha \to [\alpha]_{\mathcal{B}, \mathcal{B}}\). \[\theta^{-1}(AB) = \theta^{-1}(I_n)\]. So \(\theta^{-1}(B) \circ \theta^{-1}(A) = \text{id}_V \implies \theta^{-1}(B) = (\theta^{-1}(A))^{1}\) and so is unique. Now \(\theta(\theta^{-1}(B) \circ \theta^{-1}(A)) = \theta (\text{id}_V) \implies B \cdot A = I_n\)

Q: Definition. Let \( V,W, \mathcal{B} , \mathcal{B'} \) where \( \mathcal{B}, \mathcal{B'} \) are the respective bases. We define \(P = (p_{ij})_{ij}\) to be the \(\mathcal{B}\) to \(\mathcal{B'}\) change of basis matrix where [...]
A: \[v_j' = \sum_{i=1}^n p_{ij}v_i\] that is \(P = [ \text{id}_V]_{\mathcal{B'}, \mathcal{B} } \)

Q: Lemma. Let \(P = [\text{id}_V]_{\mathcal{B'}, \mathcal{B}}\) be the change of basis matrix for \(\mathcal{B'}, \mathcal{B}\). Then P is invertible and \(P^{-1} = [\text{id}_V]_{\mathcal{B}, \mathcal{B'}}\)
A: Let \(Q = [\text{id}_V]_{\mathcal{B}, \mathcal{B'}}\) \[PQ = [\text{id}_V]_{\mathcal{B'}, \mathcal{B}} \cdot [\text{id}_V]_{\mathcal{B}, \mathcal{B'}} =  [\text{id}_V]_{\mathcal{B},\mathcal{B'}} = I_n\] and the result follows.

Q: Theorem. Let \(\alpha : V \to W\), \(\mathcal{B}, \mathcal{B'}\) are bases for V. \(\mathcal{C}, \mathcal{C'}\) are bases for W. Let \(A = [\alpha]_{\mathcal{B}, \mathcal{C}}\) and \(A' = [\alpha]_{\mathcal{B'}, \mathcal{C'}}\). Then \[A' = Q^{-1}AP\], where \(P = [\text{id}_V]_{\mathcal{B'}, \mathcal{B}}\) and \(Q = [\text{id}_V]_{\mathcal{C'}, \mathcal{C}}\).
A: Consider first \(AP = [\alpha]_{\mathcal{B}, \mathcal{C}} \cdot [\text{id}_V]_{\mathcal{B'}, \mathcal{B}} = [\alpha]_{\mathcal{B'}, \mathcal{C}}\).
Now \(Q^{-1}AP = [\text{id}_V]_{\mathcal{C}, \mathcal{C'}} \cdot [\alpha]_{\mathcal{B'}, \mathcal{C}} = [\alpha]_{\mathcal{B'}, \mathcal{C'}}\)

Q: Definition. Let \(A,B \in \text{Mat}_{m,n}(\mathbb{F})\) we say that \(A,B\) are similar if [...]
A: there exists invertible matrices \(P,Q\) s.t. \[A = Q^{-1}BP\]

Q: Definition. Let \(A \in \text{Mat}_{m,n}(\mathbb{F})\). Then
i) Column(Row) space of A is [...]
ii) Column(Row) rank of A is [...]
A: \(<v_1 , \dots , v_n> \) where \(v_j\) is the jth column(row) of A; the dimension of the space \(\dim (<v_1 , \dots , v_n>)\)

Q: Lemma. If \(A,B \in M_{m,n}(\mathbb{F})\) and \(A \sim B\) then \(\text{rank}(A) = \text{rank}(B)\)
A: Let \(\alpha : \mathbb{F}^n \to \mathbb{F}^m\) be the linear map defined by \( x \to Ax \). Note that \[\text{rank}(\alpha)= \text{rank}(A)\]
\(B \sim A\) so we can write \(B = Q^{-1}AP\) where \(P,Q\) invertible. Let \((v_1 , \dots v_n), (w_1 , \dots , w_m)\) be the columns of P and Q respectively.
Let \(\mathcal{B} = \{e_1 , \dots , e_n\} , \mathcal{C} = \{e_1 , \dots e_m\}\) and \(\mathcal{B}' = \{v_1 , \dots , v_n\} , \mathcal{C}' = \{w_1 , \dots , w_m\}\).
The \(\mathcal{B,B'}\) change of basis matrix is \(P = [\text{id}]_{\mathcal{B',B}}\) and the \(\mathcal{C,C'}\) change of basis matrix is \(Q = [\text{id}]_{\mathcal{C',C}}\) so \(Q^{-1} = [\text{id}]_{\mathcal{C,C'}}\). Note that the \(v_i, w_i\) must span their spaces since P,Q are invertible and so the maps are isomorphism so by rank-nullity \(v_1 \dots v_n , w_1 \dots w_m\) are bases of \(\mathbb{F}^n , \mathbb{F}^m\)
So note \(B = Q^{-1}AP = [\alpha]_{\mathcal{B',C'}}\) so \(\text{rank}(\alpha) = \text{rank}(B) \implies \text{rank}(A) = \text{rank}(B)\)

Q: Theorem. Let \(A \in \text{Mat}_{m,n}(\mathbb{F})\) then \(\text{rank}(A) = \text{rank}(A^T)\). That is row rank of A = column rank of A.
A: We can write \(Q^{-1}AP\) as a matrix where the top left block is the identity and 0 otherwise. Then we can take the transpose to see \[\text{rank}(A^T)= \text{rank}(P^T A^T (Q^T)^{-1}) = r\] from how out upper left block matrix looks. Similarly \[\text{rank}(A) = \text{rank}(Q^{-1}AP) = r\].

Q: Theorem. Let \(A,B \in \text{Mat}_{m,n}(\mathbb{F})\). Then \[A \sim B \iff \text{rank}(A) = \text{rank}(B)\]
A: (->) already proved, basically just make the top left only matrix and use that
(<-) Find \(P_1 , Q_1 , P_2 , Q_2\) invertible matrices such that \(Q_1^{-1}AP_1\) and \(Q_2^{-1}BP_2\) have only top left matrix as \(I_r\) where r is the rank of A and B.  So then these matrices are the same and setting them equal we can show A is similar to B

Q: Definition. An elementary column operation is
[...]
A: i) swap columns i,j
ii)replace column i by \(\lambda\) multiplied by the column
iii) add \(\lambda\) multiplied by column i to column j
Back Extra: These can all be represented by invertible matrices.
We can use this to turn every matrix into a top left identity matrix.

Q: Lemma. \(\det A = \det A^T\)
A: Write \[\det A \sum_{\sigma} \epsilon (\sigma) \prod_{i=1}^n a_{i , \sigma(i)} = \sum_{\sigma^{-1}} \epsilon (\sigma^{-1}) \prod_{i=1}^n a_{i , \sigma^{-1}(i)} = \sum_{\sigma^{-1}} \epsilon (\sigma) \prod_{j=1}^n a_{\sigma (j) ,j}\]
And this is equal to \[\sum_{\sigma} \epsilon (\sigma) \prod_{j=1}^n a_{\sigma(j) , j} = \det A^T\]

Q: Lemma. Let A be an upper triangular matrix. Then \(\det (A) = \prod_{i=1}^n a_{i,i}\)
A: \[\det A = \sum_{\sigma \in S_n} \epsilon (\sigma) \prod_{i=1}^n a_{i , \sigma(i)}\]  Note if \(\sigma(i) < i\) then \(a_{i, \sigma(i)} = 0\) so we have all terms being 0 except for the identity permutation which gives us our result.

Q: Definition. A volume form on \(\mathbb{F}\) is a function \(d : \mathbb{F}^n \times \dots \times \mathbb{F}^n \to \mathbb{F}\) that satisfies
[...]
A: i) multi-linear \[d(v_1 , \dots , v_i + v_i' , \dots v_{i+1} , \dots v_n) = d(v_1 , \dots , v_i , \dots v_n) + d(v_1 , \dots , v_i' , \dots , v_n)\] \[d(v_1 , \dots , \lambda v_i , \dots v_n) = \lambda d(v_1 , \dots , v_n)\] fo fall \(v_1 \dots v_n v_1' \dots v_n' \in \mathbb{F}^n\) and \(\lambda \in \mathbb{F}\)
ii) Attenuating if \(v_i = v_j\) for some \(i \neq j\) then \[d(v_1 \dots v_n) = 0\]

Q: Lemma. Let \(A \in \text{Mat}_n (\mathbb{F})\) then \[(A^{(1)} \dots A^{(n)} \to \det A)\] is a volume form.
A: Pretty easy to check properties. \[\det (A^{(1)} \dots A^{(i)} + b \dots A^{(n)}) = \sum_{\sigma} \epsilon (\sigma) \left [ \prod_{j = 1, j \neq i}^n a_{\sigma (j),j}\right ](a_{\sigma(i), i} + b_{\sigma (i)}) = \det (A^{(1)} \dots A^{(n)}) + \det (A^{(1)} \dots b \dots A^{(n)})\] after expanding.
If \(A^{(j)} = A^{(k)}\), let \(\tau\) be the permutation that swaps j,k which only changes the det to a minus sign, but everything else stays the same.

Q: Definition. Let \(A \in M_n(\mathbb{F})\). We define \[\det A = [...] \]
A: \sum_{\sigma \in S_n} \epsilon (\sigma) \prod_{i=1}^n a_{i , \sigma(i)}

Q: Lemma. If d is a volume form on \(\mathbb{F}\). Then \[d(v_1 \dots v_i \dots v_j \dots v_n) = -d(v_1 \dots v_j \dots v_i \dots v_n)\]
A: \( 0 = d(v_1 \dots v_i + v_j \dots v_i + v_j \dots v_n) = d(v_1 \dots v_i \dots v_j \dots v_n) + d(v_1 \dots v_i  \dots v_i \dots v_n) + d(v_1 \dots v_j \dots v_i \dots v_n) + d(v_1 \dots v_j \dots v_j \dots v_n) \) and the middle two terms are 0 so this gives what we want

Q: Corollary. If d is a volume form on \(\mathbb{F}\) then \[d(v_{\sigma(1)} \dots v_{\sigma(n)}) = \epsilon(\sigma)d(v_1 \dots v_n)\] for any permutation \(\sigma\)
A: Write \(\sigma\) as a product of transpositions. and use the fact that we switch sign whenever we switch two things.

Q: Lemma. Let d be a volume form on \(\mathbb{F}\) then if \[A = [A^{(1)} \dots A^{(n)}] \in \text{Mat}_n(\mathbb{F})\] then \[d(A^{(1)} , \dots , A^{(n)}) = d(e_1 , \dots e_n) \cdot \det A\]
A: \(d(A^{(1)} , \dots , A^{(n)}) = d(\sum_{i=1}^n a_{i1} e_i , \dots A^{(2)} \dots A^{(n)}) = \sum_i a_{i1} d(e_i, \dots A^{(2)} \dots A^{(n)}) = \dots = \sum_{i_1 , \dots i_n} a_{i_1 1} a_{i_2 2} \dots a_{i_n n} d(e_{i_1} \dots e_{i_n}) = \sum_{\sigma} a_{\sigma(1) 1} \dots a_{\sigma(n) n} d(e_{\sigma(1)} \dots e_{\sigma(n)})\)
But now this is equal to \[d(e_1 \dots e_n) \sum_{\sigma} a_{\sigma(1) 1} \dots a_{\sigma(n) n} \epsilon(\sigma) = d(e_1 \dots e_n) \det A\]

Q: Theorem. Let \(A,B \in \text{Mat}_n (\mathbb{F})\) then \[\det AB = \det A \det B\]
A: Lets write \[B = [B^{(1)} \dots B^{(n)}]\] where those are the columns of the matrix and define \[d_A : (B^{(1)} \dots B^{(n)}) \to \det (AB)\] we can now check that \(d_A\) is a volume form. Note that \[AB = [A B^{(1)} \dots A B^{(n)}] \] so A acts linearly on each columns so \(d_A\) is a volume form. Now \[\det AB = d_A(B) = d_A(e_1 \dots e_n) \det B = \det A \det B\]

Q: Corollary. If \(A  = \text{Mat}_n (\mathbb{F})\) is invertible then \(\det A \neq 0\) and moreover \[\det A^{-1} = \frac{1}{\det A}\]
A: \(I = A A^{-1}\) \[1 = \det I = \det (A A^{-1}) = \det A \det A^{-1}\]

Q: Theorem. Let \(A \in \text{Mat}_n (\mathbb{F})\). The following are equivalent.
i) A is invertible.
ii) \(\det A \neq 0\)
iii) \(\text{rank} A = n\)
A: (i) -> (ii) is done already from determinant definitions.
(ii) -> (iii) Prove by contrapositive. Let A have rank(A)<n. Then there exists a non-zero vector v s.t. \(Av = 0\). Let \(v_i \neq 0\) be the largest non-zero entry of v. Then define
Then \(\det B \neq 0\), but \(\det AB = 0\) so \(\det A = 0\).
(iii) -> (i) We have \(\alpha : x \mapsto Ax, \alpha \mathbb{F}^n \to \mathbb{F}^n\) is surjective and linear and therefore isomorphic and if \(\mathcal{B} = \{e_1 \dots e_n\}\) is the standard basis, \[[\alpha]_{\mathcal{B} , \mathcal{B}} = A\] so \([\alpha^{-1}]_{\mathcal{B,B}} = A^{-1}\). So A is invertible.

Q: Theorem. For \(A \in \text{Mat}_n (\mathbb{F})\), \[A \cdot (\text{adj} A) = (\det A) \cdot I\] and \[(\text{adj} A )\cdot A = (\det A) \cdot I\]
A: \(((\text{adj} A) A)_{jk} = \sum_{i=1}^n (\text{adj} A)_{ji} A_{ik} = \sum (-1)^{i+j} a_{ik} \det (\hat{A_{ij}})\)
If \(k = j\) then this is equal to \(\det A\) from expansion.
If \(k \neq j\) then this is 0 as we are taking the determinant of A but replacing the jth column with the jth column.
We can prove the other direction in a similar way.

Q: Corollary. If \(\det A \neq 0\) then \[A^{-1} = \frac{\text{adj} A }{\det A}\]
A: Comes from the fact that \((\text{adj} A) A = (\det A )I\)

Q: Lemma. Let A,B,C be square matrices. Then
A: Let's call that matrix M. Then \[\det M = \sum_{\sigma} \epsilon (\sigma) \prod_{i=1}^n m_{i, \sigma(i)}\].
We claim that if \(\sigma\) is a permutation of \(1 \dots n\) s.t. \(\sigma(i) > k\) for some \(i \in \{1 \dots k\}\) then \[\prod_{i=1}^n m_{i, \sigma(i)} = 0\].
So we claim that there exists \(j \in \{k+1 , \dots , n\}\) for which \(\sigma(j) \leq k\). Just consider \(\sigma (i) \sigma^2 (i) \dots \sigma^l (i) = i\) for some l.
So there is some minimal r s.t. \(\sigma^r(i) < k\).
Otherwise this means that \(\sigma = \pi \rho\) s.t. \(\pi\) permutes \(1 \dots k\) and \(\rho\) permutes \(k+1 \dots n\).  So the determinant is \[\sum_{\pi} \sum_{\rho} \epsilon(\pi) \epsilon(\rho) \left [ \prod_{i=1}^k m_{i, \pi(i)}\right ] \left [ \prod_{i = k+1}^n m_{i, \rho(i)}\right ] = \det A \det B\] after we separate the products.

Q: Corollary. If \(A_1 \dots A_n\) are square matrices then
A: Comes from the fact that if a matrix is upper triangular and made up of 3 square matrices, we just get the det of the top left multiplied by det of bottom right.

Q: Definition. Let \(A,B \in \text{Mat}_n(\mathbb{F})\) we say that \(A,B\) are similar if [...]
A: there exists invertible P such that \[P^{-1}AP = B\]

Q: Definition. An endomorphism of a vector space V is [...]. Define \[\text{End}(V) = [...]\]
A: a linear map \(\alpha : V \to V\); \mathcal{L}(V,V)
Back Extra: Notation: \([\alpha]_{\beta} = [\alpha]_{\beta , \beta}\) where \(\alpha\) is a linear map and \(\beta\) a basis.
If \(\alpha \in \text{End}(V)\) and \(\beta , \beta '\) are bases of V then \[P^{-1} [\alpha]_{\beta} P = [\alpha]_{\beta '}\] where P is the \(\beta\) to \(\beta '\) change of basis matrix.

Q: Definition. For \(A \in \text{Mat}_n(\mathbb{F})\) define the trace of A  [...] also define the characteristic polynomial of A as [...]
A: \[\text{tr}(A)  = \sum_{i=1}^n a_{ii}\]; \[\mathcal{X}_A (t)  = \det (t \cdot I_n - A)\]

Q: Lemma. If \(A,B \in \text{Mat}_n (\mathbb{F})\) then
\(\text{tr}(AB) = \text{tr}(BA)\)
\(\det A = \det B\) and \(\text{tr}(A) = \text{tr}(B)\)whenever A is similar to B
\(\mathcal{X}_A(t) = \mathcal{X}_B(t)\) whenever A is similar to B
A: \(\text{tr}(AB) = \sum_i \sum_j a_{ji}b_{ji} = \sum_j \sum_i a_{ij}b_{ji} = \sum_j \sum_i b_{ji}a_{ij} = \text{tr}(BA)\)
\(\text{tr} A  = \text{tr}(P^{-1}BP) = \text{tr}(BPP^{-1}) = \text{tr}(B)\) when A is similar to B.
Also \(\det A = \det P^{-1} B P = \det P^{-1} \det B \det P = \det B\)
Also \(\mathcal{X}_A(t) = \det (t \cdot I - A) = \det(t \cdot I - P^{-1}BP) = \det(P^{-1}(t \cdot I - B) P) = \det(tI - B)\)

Q: Definition. Let \(\alpha \in \text{End}(V)\). We define \(\text{tr}(\alpha) = \text{tr}([\alpha]_{\beta}), \; \det \alpha = \det ([\alpha]_{\beta})\) and \(\mathcal{X}_{\alpha} (t) = \det (t \cdot \text{id}_V - \alpha)\) for \(\beta\) a basis of V.
A: These are well defined as tr, det and \(\mathcal{X}_A\) don't depend on basis.

Q: Definition. Let \(\alpha \in \text{End}(V)\)
i) \(\lambda \in \mathbb{F}\) is an eigenvalues of \(\alpha\) if [...]
ii) \(v \in V\)  is an eigenvector if [...]
A: there \(\exists v \in V\), \(v \neq 0\) and \(\alpha (v) = \lambda \cdot ve\); \(v \neq 0\) and \(\alpha (v) = \lambda v\) for some \(\lambda \in F\)

Q: Definition. Say that \(\alpha \in \text{End}(V)\) is diagonizable if [...]
A: there exists a basis \(\beta\) of V s.t. \([\alpha]_{\beta}\) is a diagonal matrix.

Q: Definition. \(\alpha \in \text{End}(V)\), the \(\lambda\)-eigenspace of \(\alpha\) is [...]
A: \[\{v : \alpha(v) = \lambda v\} = \text{Ker}(\text{id}_V \cdot \lambda - \alpha) \]
Back Extra: This is a vector space.

Q: Lemma. Let \(\alpha \in \text{End}(V)\), let \(\lambda_1 , \dots , \lambda_k \in \mathbb{F}\) be the eigenvalues of \(\alpha\). Then \(V_{\lambda_1} + \dots + V_{\lambda_k}\) is a direct sum.
A: Let \(x_i \in V_{\lambda_i}\) and \(y_i \in V_{\lambda_i}\) be such that \[x_1 + \dots + x_k = y_1 + \dots y_k\]
We consider the map \(\beta_i = \prod_{j \neq i, j=1}^n (\lambda_j \cdot \text{id}_V - \alpha)\)
Let \(v_l = x_l - y_l\) for \(l = 1, \dots , k\).
Note that \(\beta_i v_l = 0\) if \(i \neq l\) and \(\prod_{j \neq l}(\lambda_j - \lambda_l) \neq 0\) if \(i = l\).
So we have \(v_1 + \dots + v_k = 0\), apply \(\beta_i\) to the sum to get \[v_i = 0 \implies x_i = y_i\]

Q: Theorem. Let \(\alpha \in \text{End}_V\) and let \(\lambda_1 \dots \lambda_k \in \mathbb{F}\) be distinct eigenvalues. The following are equivalent.
i) \(\alpha\) is diagonalisable
ii) V has a basis consisting of eigenvectors of \(\alpha\)
iii) \(\sum_{i=1}^k \dim(V_{\lambda_i}) = \dim V\)
A: (i) -> (ii) : Let \(\beta\) be a basis for V for which \([\alpha]_{\beta}\) is diagonal.
This means that \(\alpha (v_i) = \mu_i v_i\) where \(\beta = \{v_1 \dots v_n\}\). Thus \(\beta\) is a basis for V of eigenvectors of \(\alpha\).
(ii) -> (i): Let \(\beta\) be a basis of eigenvectors. Note \([\alpha]_{beta}  = \text{diag} (\lambda_1 \dots \lambda_n)\) where \(\lambda_i\) are defined as \(\alpha (v_i) = \lambda_i v_i\).
(ii) <-> (iii): \((ii) \iff V_{\lambda_1} + \dots + V_{\lambda_k} = V\) with forward direction coming from the fact that each \(v_i\) in the basis is in some \(V_{\lambda_i}\). The backward direction comes from building up a basis using SEL. Now this is true iff \(V_{\lambda_1} \bigoplus \dots \bigoplus V_{\lambda_k} = V \iff \dim(V_{\lambda_1} \bigoplus \dots \bigoplus V_{\lambda_k}) = \dim V \).

Q: Definition. Let \(\alpha\) be an endomorphism, we say it is triangularizable if [...]
A: there exists a basis \(\beta\) of V s.t. \([\alpha]_{\beta}\) is upper triangular.

Q: Definition. The minimal polynomial of \(\alpha \in \text{End}(V)\) is [...]
A: the monic polynomial \(m_{\alpha} \in \mathbb{F}(t)\) with minimum degree such that \[m_{\alpha}(\alpha) = 0\].

Q: Lemma. Let \(\alpha \in \text{End}(V), p(t) \in \mathbb{F}(t)\). Then;
i) \(m_{\alpha}\) always exists
ii) \(p(\alpha) = 0 \iff m_\alpha | p\)
iii) the minimal polynomial is unique
A: i) Note that \(\text{id}_V , \alpha , \dots , \alpha^{n^2} \in \text{End}(V)\) and \(\dim(\text{End}(V)) = n^2\) so there must exist some combination that gives 0.
ii) Apply division algorithm to write \(p(t) = q(t)m_{\alpha}(t) + r(t)\) with degree r less than degree of minimal polynomial. So \(r(\alpha)= 0\) so we must have r = 0.
iii) If \(m_{\alpha}\) and \(m_{\alpha}'\) are minimal polynomials then they divide each other so \(m_{\alpha} = c m_{\alpha}'\)and it is monic so c = 1.

Q: Lemma. Let \(\alpha \in \text{End}(V), \lambda \in \mathbb{F}\). TFAE:
i) \(\lambda\) is an eigenvalues of \(\alpha\)
ii) \(\mathcal{X}_{\alpha}(\lambda) = 0\)
iii) \(m_{\alpha}(\lambda) = 0\)
A: (i) <-> (ii) already proven.
(iii) -> (ii) Cayley Hamilton says that \(\mathcal{X}_{\alpha}(\alpha) = 0\), so \(m_{\alpha} | \mathcal{X}_{\alpha}\). So \(m_{\alpha}(\lambda) = 0 \implies \mathcal{X}_{\alpha}(\lambda) = 0\)
(ii) -> (iii) Write \(m_{\alpha}(\alpha) = \sum_{i=0}^d c_i \alpha^i\) so let v be an eigenvector with eigenvalue \(\lambda\). We have \[0 = m_{\alpha}(\alpha)v = (\sum_{i=0}^d c_i \alpha^i)v\] so \(m_{\alpha}( \lambda) = 0\).

Q: Definition. Let \(\alpha \in \text{End}_V\), \(\lambda \in \mathbb{F}\) be eigenvalue of \(\alpha\). The algebraic multiplicity of \(\lambda\) is [...] The geometric multiplicity of \(\lambda\) is [...] Denote the AM \(a_{\lambda}\) and GM \(g_{\lambda}\) and also let \(c_{\lambda}\) denote the multiplicity of \(\lambda\) as a root of \(m_{\alpha}\).
A: the multiplicity of \(\lambda\) as a root of \(\mathcal{X}_{\alpha}(t)\).; \[\dim V_{\lambda} = \dim (\text{Ker}(\lambda \text{id}_V - \alpha))\]

Q: Lemma. Let \(\lambda \in \mathbb{F}\) be an eigenvalue of \(\alpha\). Then
i) \(1 \leq g_{\lambda} \leq a_{\lambda}\)
ii) \(1 \leq c_{\lambda} \leq a_{\lambda}\)
A: We know that \(g_{\lambda}, c_{\lambda} \geq 1\). Now to prove (i). Let \(V_{\lambda}\) be the \(\lambda\)-eigenspace. Let \(\mathcal{B}_{\lambda} = \{v_1 \dots v_{g_{\lambda}}\}\)be a basis for \(V_{\lambda}\) and let \(\beta = \{v_1 \dots v_n\}\) and write
So \(\mathcal{X}_{\alpha}(t) = \det ((t-\lambda)I) \cdot \det (t \cdot I - C) = (t-\lambda)^{g_{\lambda}} \det (t \cdot I - C)\)
So \(a_{\lambda} \geq g_{\lambda}\).
ii)Use C-H to see that \(m_{\alpha} | \mathcal{X}_{\alpha}(t)\) so \(c_{\lambda} \geq a_{\lambda}\) as \((t-\lambda)^{c_{\lambda}} | m_{\alpha}(t)\) implies it divides the characteristic polynomial too.

Q: Lemma. Take \(\mathbb{F} = \mathbb{C}\). The following are equivalent.
i) \(\alpha \in \text{End}(V)\) is diagonisable
ii) \(g_{\lambda} = a_{\lambda}\) for all eigenvalues
iii) \(c_{\lambda} = 1\)
A: Exercise

Q: Definition. \(A \in \text{Mat}_n(\mathbb{C})\) is in Jordan Normal Form [...]
where \(J_{n_i}(\lambda_i)\) are Jordan Blocks defined by
[...]
A: if it is in block-diagonal form; so \(J_n(\lambda) = \lambda \cdot I + J_n(0)\)

Q: Theorem. Every matrix \(A \in \text{Mat}_n(\mathbb{C})\) is similar to a matrix in Jordan Normal Form.
A: bruh

Q: Definition. Let V be a vector space over a field \(\mathbb{F}\). The dual space is [...]
A: \[V^* = \{\theta \text{ linear map} : \theta : V \to \mathbb{F}\} = L(V , \mathbb{F})\]
Back Extra: This is a vector space

Q: Lemma. Let V be a vector space with basis \(\beta = \{v_1 , \dots ,v_n\}\). Define the linear function as \(\{\epsilon_1 , \dots , \epsilon_n\}\) by \[\epsilon_i(v_j) = \delta_{ij}\] then \(\{\epsilon_1 , \dots , \epsilon_n\}\) is a basis for \(V^*\)
A: Assume that \[\sum_{i=1}^n c_i \epsilon_i = 0 \in V^*\] we have \[0 = (\sum_{i=1}^n c_i \epsilon_i)(v_j) = \sum_{i=1}^n c_i \epsilon_i (v_j) = c_j\] so all coefficients are 0 and these \(\epsilon_i\) are all L.I.
Let \(\tilde{\theta} = \sum_{i=1}^n \theta (v_i) \cdot \epsilon_i\). Note \[\tilde{\theta}(v_j) = \sum_{i=1}^n \theta(v_i)\epsilon_i (v_j) = \theta(v_j)\] so we span and therefore this is a basis.

Q: Definition. [...] is known as the dual basis.
A: \(\{\epsilon_1 , \dots , \epsilon_n\}\)
Back Extra: Let \(x = \sum_{i=1}^n c_i v_i \in V\), and \(\theta = \sum_{i=1}^n b_i \epsilon_i \in V^*\). Then \[\theta(x) = \sum_i c_i \theta(v_i) = \sum_i c_i (\sum_j b_j \epsilon_j)(v_i) = \sum_i c_i b_i\] which is the dot product of b and c. So we may think of the points of V as column vectors, and the elements of V* as row vectors which act on V via taking inner products.

Q: Corollary. \[\dim V = \dim V^*\]
A: asdfasdf

Q: Definition. \(U \leq V\) then the annihilator of U is [...]. Likewise, we define, if \(U \leq V^*\) [...]
A: \[V^* \geq U^0 = \{\theta \in V^* : \theta(u) = 0 \; \forall u \in U\}\]; \[U^0 = \{v \in V : \theta(v) = 0 \; \forall \theta \in U\} = \bigcap_{\theta \in U} \text{Ker}(\theta)\]

Q: Lemma. Let V be a finite dimensional space. Let \(U \leq V\) then \[\dim U + \dim U^0 = \dim V\]
A: Let \(u_1 \dots u_k\) be a basis for U and extend to a basis for V, \(u_1 \dots u_n\). Let \(\epsilon_1 , \dots , \epsilon_n\) be the dual basis to \(u_1 \dots u_n\). Note that \(<\epsilon_{k+1} \dots \epsilon_n> \subseteq U^0\) since \(\epsilon_j (u_i) = 0\) for all \(j \geq k+1 , i < k\), so each \(\epsilon_j \; j \geq k+1\) maps each basis vector of U to zero so maps all of U to zero.
Let \(\theta \in U^0\) and \(\theta = \sum_{i=1}^n c_i \epsilon_i\). Now note for \(i \leq k\) \[0 = \theta(u_i) = (\sum_{j=1}^n c_j \epsilon_j)(u_i) = \sum_{j=1}^n c_j \epsilon_j (u_i) = c_i\] so \[\theta = \sum_{i = k+1}^n c_i \epsilon_i \in < \epsilon_{k+1} \dots \epsilon_n >\]

Q: Definition. Let \(\alpha \in L(V,W)\), where V,W are vector spaces over \(\mathbb{F}\). The dual map \(\alpha^* : W^* \to V^*\) is defined by [...]
A: \[\theta \mapsto \theta \circ \alpha\]
Back Extra: This is a linear functional, easy to check its linear and domains and ranges are correct. It is a linear map between W* and V*.

Q: Lemma. Let \(\alpha \in L(V,W)\), where V,W are finite dimensional vector spaces over \(\mathbb{F}\). Let \(\mathcal{B,C}\) be bases of V,W. Now \(\mathcal{B^*, C^*}\) be the dual bases for the dual spaces V*, W*. Then \[[\alpha^*]_{\mathcal{C^*, B^*}} = ([\alpha]_{\mathcal{B,C}})^T\]
A: Let \(\mathcal{B} = \{\beta_1 \dots \beta_n\}\), \(\mathcal{C} = \{\gamma_1 \dots \gamma_m\}\) and \(\mathcal{B^*} = \{\beta_1^* \dots \beta_n^*\}, \mathcal{C^*} = \{\gamma_1^* \dots \gamma_m^*\}\) also let \(A = [\alpha]_{\mathcal{B,C}} = (a_{ij})\) and \(A^* = [\alpha^*]_{\mathcal{C^*, B^*}} = (a_{ij}^*)\) \[\alpha(\beta_k) = \sum_{i=1}^m a_{ij}\gamma_i\] \[\alpha^* (\gamma_j^*) = \sum_{i=1}^n a_{ij}^* \beta_i^* \in V^*\] so \[\alpha^* (\gamma_j^*) (\beta_k) = \sum_{i=1}^n a_{ij}^* \beta_i^* (\beta_k) = a_{kj}^*\] But also \[\alpha^* (\gamma_j^*)(\beta_k) = (\gamma_j^* \circ \alpha)(\beta_k) = \gamma_j^* (\alpha(\beta_k)) = \gamma_j^* (\sum_{i=1}^m a_{ik}\gamma_i) = \sum_{i=1}^m a_{im}\gamma_i^* (\gamma_i) = a_{jk}\] Therefore \[a_{kj}^* = a_{jk}\] so \(A = (A^*)^T\)

Q: Lemma. Let \(\alpha \in L(V,W)\), V,W be finite dimensional. Then
i) \(\text{Ker}(\alpha^*) = (\text{Im} (\alpha))^0\)
ii) \(\text{rank}(\alpha^*) = \text{rank}(\alpha)\)
iii) \(\text{Im}(\alpha^*) = (\text{Ker}( \alpha))^0 \)
A: i) Let \(\theta \in \text{Ker}(\alpha^*) \iff \alpha^* (\theta) = 0 \iff \theta \circ \alpha = 0 \iff \theta \in (\text{Im}(\alpha))^0\)
ii) \(\text{rank}(\alpha^*) = \text{rank}([\alpha^*]_{\mathcal{B' , C'}}) = \text{rank}([\alpha]_{\mathcal{C,B}})\)
iii) Let \(\theta \in \text{Im}(\alpha^*) \subseteq V^*\). Thus \(\exists \psi \in W^*\) s.t. \(\alpha^* (\psi) = \theta\)
\(\theta = \psi \circ \alpha\). Note \(\forall v \in \text{Ker}(\alpha)\) we have \[\theta(v) = (\psi \circ \alpha)(v) = \psi(\alpha(v)) = 0\]
\(\theta \in (\text{Ker} \alpha)^0\). So \(\text{Im}(\alpha^*) \subseteq (\text{Ker} \alpha)^0\)
Now note \[\dim (\text{Ker}(\alpha)) + \dim (\text{Ker}\alpha)^0 = \dim V\]
\[\dim(\text{Ker}(\alpha)) + \dim(\text{Im}\alpha) = \dim V\]
So we have \[\dim (\text{Ker}\alpha)^0 = \dim(\text{Im}\alpha) = \text{rank}(\alpha) = \text{rank}(\alpha^*) = \dim(\text{im}(\alpha^*))\] and so the dimensions are the same so we must have equality.

Q: Definition. Let \(V,W\) be vector spaces over \(\mathbb{F}\). We say \(\psi : V \times W \to \mathbb{F}\) is a bilinear form if [...]
A: it is linear in each argument.

Q: Definition. Let V be a finite dimensional vector space over \(\mathbb{F}\) with a basis \(\beta = \{v_1 \dots v_n\}\), we resent the linear form \(\psi : V \times V \to \mathbb{F}\) by the matrix [...]
A: \[[\psi]_{\beta} = (\psi(v_i,v_j))_{i,j = 1}^n\]
Back Extra: \(\psi\) is determined by \([\psi]_{\beta}\)

Q: Lemma. Let \(\beta = \{v_1 \dots v_n\}, \beta' = \{v_1' \dots v_n'\}\), let P be the \(\beta\) to \(\beta'\) change of basis matrix. Then \[[\psi]_{\beta'} = [...]\]
A: P^T [\psi]_{\beta} P

Q: Lemma. We say \(A,B \in \text{Mat}_n(\mathbb{F})\) are congruent if [...]
A: there exists \(P \in \text{Mat}_n(\mathbb{F})\) such that \[A = P^TBP\]

Q: Definition. A bilinear form \(\psi : V \times V \to \mathbb{F}\) is called symmetric if [...]
A: \[\psi(x,y) = \psi(y,x)\] for all \(x,y \in V\)

Q: Lemma. Polarisation Identity. If \(\psi : V \times V \to \mathbb{F}\) is symmetric then \(\psi\) is determined uniquely by its diagonal \(\psi(u,u)\) for all \(u \in U_0\).
A: For all \(u,v \in V\) \[\psi(u,v) = \frac{1}{2}(\psi(u+v,u+v) - \psi(u,u) - \psi(v,v))\]

Q: Definition. We say \(Q : V \to \mathbb{F}\) is a quadratic form if:
[...]
A: i) \(Q(\lambda v) = \lambda^2Q(v) \; \; \forall \lambda \in \mathbb{F}, \forall v \in V\)
ii) The map \((u,v) \mapsto Q(u+v) - Q(u) - Q(v)\) is bilinear.
Back Extra: Idea here is that \(Q(u) = \psi(u,u)\) is a quadratic form.

Q: Definition. If \(U \leq V\) we define \[U^{\perp} = [...]\]
A: \{v \in V : \psi (u,v) = 0 \; \forall u \in U\}

Q: Theorem. Let V be a finite dimensional vector space. Let \(\psi : V \times V \to \mathbb{F}\) a symmetric bilinear form. Then there exists a basis \(\beta\) such that \([\psi]_{\beta}\) is diagonal.
A: We work by induction on \(n = \dim (V)\). Let \(w \in V\) s.t. \(Q(w) = \psi(u,u) \neq 0\). Such a direction must exists else we are done by polarisation identity and we have \(\psi = 0\).
We claim that \(V = <w> \bigoplus <w>^{\perp}\). Let \(v \in V\) and write \[v = \lambda w + (v - \lambda w)\] for some \(\lambda \in \mathbb{F}\) chosen so that \(v - \lambda w \in <w>^{\perp}\)
\[\psi(w, v - \lambda w) = \psi(w,v) - \lambda \psi(w,w)\] so choose \(\lambda = \frac{\psi(w,v)}{\psi(w,w)}\).
Also \(<w> \cap <w>^{\perp} = \{0\}\). To see this let \(v \in <w> \cap <w>^{\perp} \implies v = \lambda w\), so \(0 = \psi(w, \lambda w) = \lambda \psi(w,w) \implies \lambda = 0\) as the other thing isn't 0 so v = 0.
Now let \(\psi'\) be the restriction of \(\psi\) to the space \[W = <w>^{\perp}\] Since direct sum \(\dim <W>^{\perp} < \dim V\) so apply induction to find a basis \(\beta' = \{v_1 \dots v_{n-1}\}\) then \(\beta = \{v_1 \dots v_n\}\) where \(v_n = w\) is a basis for V where

Q: Corollary. Let C be a vector space over \(\mathbb{R}\). Let \(\psi : V \times V \to \mathbb{R}\) be a symmetric bilinear form. Then \(\exists\) basis \(\beta\) such that
A: First apply our theorem to get a basis \(\beta'\) for which we have
and now just redefine \[v_i = \frac{v_i'}{\sqrt{|\lambda_i|}}\] for \(i \leq p+q\)
where \(\beta' = \{v_1' \dots v_n'\}\). So \[\psi(v_i,v_i) = \frac{1}{|\lambda_i|} \psi(v_i', v_i') \in \{\pm 1\}\]

Q: Definition. We call p+q the [...] and p-q the [...], where we have written the symmetric bilinear form with a positive and negative identity matrix on the diagonal.
A: rank of \(\psi\); signature

Q: Definition. Let Q be a quadratic form on vector space V over \(\mathbb{R}\). Let \(U \leq V\).
We say that U is positive definite if [...]
We say U is positive semi definite if [...]
Same definition for negative definite and negative semi definite.
A: \(Q(u)>0, u \in U \setminus \{0\}\); \(Q(u) \geq 0, \forall u \in U\)

Q: Theorem. Sylvester's Law of Inertia. Let V be a vector space over \(\mathbb{R}\). Let \(\psi : V \times V \to \mathbb{R}\) be bilinear. Then the signature of \(\psi\) is well defined. In other words if \(\beta, \beta'\) are basis of V and
then p = p', q = q'.
A: Let \(\beta = \{v_1 \dots v_n\}, \beta' = \{v_1' \dots v_n'\}\). \(P = <v_1 \dots v_p>, N = <v_{p+1} \dots v_n>\) and let \(\hat{N} = <v_{p'+1}' \dots v_n'>\)/ Now note that for \(Q(u) = \psi(u,u)\), P is positive definite, \(N, \hat{N}\) are negative semi definitive. Now note \[\dim P + \dim N = \dim V\]
Also \[\dim V \geq \dim(P + \hat{N}) = \dim(P \bigoplus \hat{N}) = \dim P + \dim \hat{N}\] as \(P \cap \hat{N} = \{0\}\) as otherwise we have \(Q(v) >0, Q(v) \leq 0\). So \(\dim \hat{N} \leq \dim N \implies p \leq p'\). Apply the same argument with the bases reversed to get \( p = p'\). We also know \[p + q = \text{rank}\psi = p' + q'\]

Q: Corollary. If I have \(A,B \in \text{Mat}_n (\mathbb{R})\) symmetric then A,B are congruent \(\iff\) A,B have the same rank and signature.
A: Follows from Sylvester's Law.

Q: Proposition. Let V be a finite dimensional vector space over \(\mathbb{C}\). Let \(\psi : V \times V \to \mathbb{C}\) be symmetric and bilinear. There exists a basis \(\beta\) of V s.t.
A: Use the theorem to put into diagonal form and then renormalise. Since \(\sqrt(-1) = i \in \mathbb{C}\) we can get rid of the negative terms.

Q: Corollary. Complex symmetric matrices are congruent \(\iff\) they have the same rank
A: Comes from the fact we can always turn a symmetric bilinear matrix over the complex numbers to just the identity in the top left block.

Q: Definition. Let V be a vector space over \(\mathbb{C}\). \(\psi : V \times V \to \mathbb{C}\) is Hermitian if:
[...]
A: (i) For all fixed \(v \in V\) the map \[u \mapsto \psi(u,v)\] is linear.
(ii) \(\psi(u,v)^* = \psi(v,u)\)
Back Extra: A Hermitian form is not necessarily bilinear but it is sesquilinear: \[\psi(\lambda u + \mu w, v) = \lambda\psi(u,v) + \mu \psi(w,v)\]
\[\psi(u, \lambda w + \mu v) = \lambda^* \psi(u,w) + \mu^* \psi(u,v)\]

Q: Proposition. Polarisation Idenitity for Hermitian Forms.\[\phi(u,v) = \frac{1}{4}[Q(u+v) - Q(u-v)+iQ(u+iv) - iQ(u-iv)]\]
A: Just use definitions and go through algebra.

Q: Lemma. Let \(B = (v_1 \dots v_m)\) be a basis of V and \(C = (w_1 \dots w_n)\) be a basis of W. Let \([\phi]_{B,C} = (\phi(v_i , w_j))\). Then, \[\phi(v,w) = [v]_B^T [\phi]_{B,C} \overline{[w]_C}\]
A: Let B,B' be bases of V and C,C' be basis of W. Let \(P = [I]_{B' , B}\) and \(Q = [I]_{C',C}\). Then \[[\phi]_{B',C'} = P^T [\phi]_{B,C} \overline{Q}\]

Q: Theorem. Let V be a finite-dimensional \(\mathbb{C}\)-vector space. Let \(\phi : V \times V \to \mathbb{C}\) be a Hermitian form on V. Then there exists a basis \(B = (v_1 , \dots , v_n)\) of V such that
\[[\phi]_B = \left ( \begin{matrix} I_p & 0 & 0 \\ 0 & -I_q & 0 \\ 0 & 0 & 0\end{matrix}\right )\]where p,w depend only on \(\phi\) and not B
A: We can repeat the same method as for the real symmetric bilinear form. If \(\phi = 0\), we are done. Otherwise use the polarisation identity to get some \(e_1\) s.t. \(\phi(e_1 , e_1) \neq 0\). Let \[v_1 = \frac{e_1}{\sqrt{|\phi(e_1 , e_1)|}} \implies \phi(v_1 , v_1) = \pm 1\]Consider the orthogonal space \(W = \{w \in V : \phi(v_1 , w) = 0\}\). Then we can check that \(V = <v_1> \bigoplus W\). Hence we can inductively diagonalise \(\phi\).
To show uniqueness we can prove that p is the maximal dimension of a subpsace on which \(\phi\) is positive definite (which is well-defined since \(\phi(u,u) \in \mathbb{R}\)). The geomtric interpretation of q is similar.

Q: Theorem. Let \(\psi: V \times V \to \mathbb{F}\) be a symmetric bilinear form or a Hermitian form. Let \(\beta = \{v_1 , \dots , v_n\}\) be a basis for V. Let \(A = [\psi]_{\beta}\). Then \(\psi\) is positive definite \(\iff \)all top left submatrices of A have positive determinant.
A: (->) Already proved.
(<-) Proof by induction on \(\dim V = n\). If n = 1 this is clearly true. Now for the basis step let's define \(U_k = <v_1 \dots v_k>\). Let's write
By considering the restriction of \(\psi\) to \(U_{n-1}\) we use induction to see that actually
for some \(c \in \{0, \pm 1\}\). So \(\det A > 0 \implies |\det P|^2 \cdot c > 0 \implies c = 1\)
So A is positive definite and therefore \(\psi\) is positive definite.

Q: Definition. An inner product is a function \(\langle \cdot, \cdot \rangle : V \times V \to \mathbb{F}\) which [...]
A: is a symmetric bilinear form or Hermitian and is positive definite.
Back Extra: A vector space together with an inner product is called an inner product space. An inner product space is a vector space when I have a notion of length and angle.

Q: Definition. For \((V, \langle . \rangle)\) we define a norm by [...]
for all \(v \in V\)
A: \[||v|| = \sqrt{\langle v , v \rangle}\]

Q: Lemma. Cauchy-Schwartz. \(\forall u,v \in V\) \[|\langle u,v \rangle | \leq ||u || ||v||\]
A: \[0 \leq \langle u - \lambda v, u - \lambda v \rangle = ||u||^2 -\lambda \langle u , v \rangle - \overline{\lambda \langle u,v \rangle} + |\lambda^2 ||v||^2\]
Choose \(\lambda = \frac{\overline{\langle u,v \rangle}}{||v||^2}\) to get the result.

Q: Lemma. Triangle Inequality. \(\forall u,v \in V\) we have \[||u + v|| \leq ||u|| + ||v||\]
A: Apply Cauchy-Schwartz to \[||u+v||^2\] to get the result.

Q: Definition. We say that vectors \(v_1 \dots v_k \in V\) are orthogonal if [...] We say that \(v_1 \dots v_k \in V\) are orthonormal if [...]
A: \(\langle v_i , v_j \rangle = 0\) for \(i \neq j\).; \(\langle v_i , v_j \rangle = \delta_{ij}\)

Q: Lemma. If \(\dim V = n < \infty\) and \(v_1 \dots v_n \in V\) are orthonormal \(\implies v_1 \dots v_n\) form a basis for V. Moreover for all \(v \in V\) we can write \[v = \sum_{i=1}^n \langle v, v_i \rangle v_i\]
A: Show linear independence. For the second part let \(x = v - \sum_{i=1}^n \langle v, v_i \rangle v_i\) and show it is 0. Use the fact that \(v_1 \dots v_n\) are linearly independent so if \(x \neq 0\) then \(\{v_1 \dots v_n x\}\) is a linear independent set which is not possible.

Q: Theorem. Gram-Schmidt Orthogonalisation. If V an inner product space with basis \(\{v_1 \dots v_n\}\) then \(\exists\) an orthonormal basis \(e_1 \dots e_n\) s.t. \(\langle e_1 \dots e_k \rangle = \langle v_1 \dots v_k \rangle\) for all \(k = 1,\dots , n\)
A: We work by induction on k. For k = 1, \(e_1 = \frac{v_1}{||v_1||}\). For \(k > 1\) assume that we have \(\langle e_1 \dots e_i \rangle = \langle v_1 \dots v_i \rangle\) \(\forall i \leq k\).
Set \(f_{k+1} = v_{k+1} - \sum_{i=1}^k \langle v_{k+1}, e_i \rangle e_i\). Now for \(j \leq k\) we have \[\langle f_{k+1}, e_j \rangle = 0\] so now just put \(e_{k+1} = \frac{f_{k+1}}{||f_{k+1} ||}\). Note that \(f_{k+1} \neq 0\) since \(e_1 \dots e_k v_{k+1}\) are linearly independent.
So by induction we are done.

Q: Proposition. If \(U \leq V\), V finite dimensional, then \[V = U \bigoplus U^{\perp}\]
A: Note that \(U \cap U^{\perp} = \{0\}\) since \(x \in U \cap U^{\perp} \implies \langle x,x \rangle = 0 \implies x = 0\). Now to show \(\dim U \bigoplus U^{\perp} = \dim V\). For this define \(\alpha(v) = (\langle e_1 , v \rangle , \dots , \langle e_k , v \rangle)\) where \(e_1 \dots e_k\) is an orthonormal basis for U so \(\alpha : V \to \mathbb{F}^k\). Note that \(\text{rank} (\alpha) = k\) therefore \(\text{null}(\alpha) = \dim V - k\). But notice \(\text{Ker}(\alpha) = U^{\perp}\), since \(\alpha(v) = 0 \iff \langle v, e_i \rangle = 0 \forall i \iff v \in U^{\perp}\).
So \(\dim (U \bigoplus U^{\perp}) = \dim U + \dim U^{\perp} = k + \dim V - k = \dim V\)

Q: Definition. If \(W = U^{\perp}\) then the projection map \(\pi (u+w) = u\) is [...]
A: the orthogonal projection onto U.

Q: Proposition. If \(U \leq V\) with orthonormal basis \(e_1 \dots e_k\) and \(\pi : V \to U\) is the orthogonal projection onto U then
(i) \(\pi(v) = \sum_{i=1}^k \langle v , e_i \rangle. \cdot e_i\)
(ii) \(||v - \pi(u)|| \leq ||v - u||\) for all \(u \in U\) and equality holds iff \(u = \pi(u)\)
A: (i) Write \(v = u+w\) where \(u \in U, w \in U^{\perp}\) \[\hat{u} = \sum_{i=1}^n \langle v , e_i \rangle e_i\]
Consider for each \(i \leq j \leq k\) \[\langle v - \hat{u}, e_j \rangle = \langle v, e_j \rangle - \sum_{i=1}^n \langle v , e_i \rangle \langle e_i, e_j \rangle = 0\]
So \(v - \hat{u} \in U^{\perp}\). So note \[V = \hat{u} + (v - \hat{u})\] since this decomposition is unique, \(u = \hat{u}\)
(ii) We note \(v - \pi(v) \in U^{\perp}, u - \pi(v) \in U\) \[ ||u - v||^2 = ||(u - \pi(v)) + (\pi(v) - v) ||^2 = ||u - \pi(v)||^2 + ||\pi(v) - v||^2 \geq ||\pi(v) - v||^2\] as the cross terms cancel, equality holds iff \(u = \pi(v)\).

Q: Proposition. Let \(\alpha \in \text{End}(V)\), V finite dimensional norm space, then there exists a unique \(\alpha^* \in \text{End}(V)\) s.t. \[\langle \alpha(u), v \rangle = \langle u, \alpha^*(v) \rangle\] for all \(u,v \in V\). Moreover if \(\mathcal{B}= \{e_1 \dots e_n\}\) is an orthonormal basis of V then \[[\alpha^*]_{\mathcal{B}} = [\alpha]_{\mathcal{B}}^{\dagger}\]
A: Let \((a_{ij}) = [\alpha]_{\mathcal{B}}, (b_{ij} = [\beta]_{\mathcal{B}})\), and we will "solve for the \(b_{ij}\) " to show that we can ensure \(\alpha^* = \beta\). \[\langle \alpha(v), u \rangle = \langle v , \beta(u) \rangle\] for all \(u,v \in V\) \[\iff \langle \alpha(e_j), e_k \rangle = \langle e_j , \beta(e_k) \rangle\] for all \(j,k = 1,\dots , n\) \[\iff \langle \sum_{i=1}^n a_{ij}e_i , e_k \rangle = \langle e_j, \sum_{i'=1}^n b_{i'k}e_{i'} \rangle \iff a_{kj} = \overline{b_{jk}}\] so we're done if we take \[[\beta]_{\mathcal{B}} = [\alpha]_{\mathcal{B}}^{\dagger}\]

Q: Definition. \(\alpha^*\) is called the adjoint of \(\alpha\) when [...]
A: \[\langle \alpha(u), v \rangle = \langle u , \alpha^*(v) \rangle\] for all \(u,v \in V\)

Q: Lemma. Let \(\alpha , \beta \in \text{End}(V)\). Then
(i)\((\alpha + \beta)^* = \alpha^* + \beta^*\)
(ii)\((\lambda \alpha)^* = \overline{\lambda} \alpha^*\)
(iii) \(\alpha^{**} = \alpha\)
(iv) \((\text{id}_V)^* = \text{id}_V\)
A: These follow from properties of the matrices \(A, A^{\dagger}\)

Q: Definition. \(\alpha \in \text{End}(V)\) we say that \(\alpha\) is self-adjoint if [...]
A: \[\alpha = \alpha^*\]
Back Extra: If a map is self adjoint and we have an orthonormal basis of V, then the matrix in this basis is symmetric or Hermitian depending on our field.

Q: Lemma. \(\alpha \in \text{End}(V)\), self-adjoint
(i) all eigenvalues of \(\alpha\) are real
(ii) eigenvectors with distinct eigenvalues are orthogonal
A: (i)Let's assume \(\mathbb{F} = \mathbb{C}\) first. Let \(\alpha(v) = \lambda v\) for some \(\lambda \in \mathbb{C}\) and \(V \in V \setminus \{0\}\). \[\langle \alpha(v) , v \rangle = \langle v , \alpha(v) \rangle \implies \lambda \langle v , v \rangle = \overline{\lambda} \langle v , v \rangle \implies \lambda = \overline{\lambda} \implies \lambda \in \mathbb{R}\]
(ii) Let \(\alpha(u) = \lambda u\), \(\alpha(v) = \mu v\) where \(\lambda \neq \mu\). \[\langle \alpha(v) , u \rangle = \langle v , \alpha (u) \rangle \implies \lambda \langle v, u \rangle = \overline{\mu} \langle v , u \rangle\] and \(\mu = \overline{\mu}\) so we have \(\langle u,v \rangle = 0\).
To extend them to \(\mathbb{R}\) just note that \(\alpha\) self adjoint over \(\mathbb{R}\) means that \([\alpha]_{\beta}\) is symmetric so it's Hermitian over \(\mathbb{C}\) so it is self-adjoint and then apply the above arguments.

Q: Theorem. Spectral Theorem for Self-Adjoint Maps. Let V be a finite dimensional inner product space. Let \(\alpha \in L(V)\) be self-adjoint. Then V has an orthonormal basis of eigenvectors of \(\alpha\). Hence it is diagonalisable in an orthonormal basis.
A: We will carry out induction on the dimension of V. Suppose \(A = [\alpha]_{B}\) with respect to the fundamental basis B. By the fundamental theorem of algebra, \(\mathcal{X}_A(\lambda)\) has a root, but since this is an eigenvalue of a self-adjoint map it must be real. Now we choose an eigenvector \(v_1 = V \setminus \{0\}\) such that \(\alpha (v_1) = \lambda v_1\). We can set \(||v_1|| = 1\) by linearity. Let \(U = \langle v_1 \rangle^{\perp} \leq V\). We then observe that \(\alpha (U) \leq U\). Indeed, let \(u \in U\). Then \(\langle \alpha(u), v_1 \rangle = \lambda \langle u, v_1 \rangle = 0\) by orthogonality. Hence \(\alpha(u) \in U\). We can then restrict \(\alpha\) to the domain U, and by induction we can the choose an orthonormal basis of eigenvectors for U. Since \(V = \langle v_1 \rangle^{\perp} \bigoplus U\) we have an orthonormal basis of eigenvectors for V.

Q: Corollary. Let V be a finite dimensional real (or complex) inner product space. Let \(\phi : V \times V \to F\) be a symmetric(or Hermitian) bilinear form. Then there exists an orthonormal basis B of V such that \([\phi]_B\) is diagonal.
A: \(A^{\dagger} = A\) hence there exists a unitary matrix P such that \(P^{-1}AP\) is diagonal. Let \(v_i\) be the ith row of \(P^{-1}\). Then \(v_1 \dots v_n\) is an orthonormal basis B of V such that \([\phi]_V\) is this diagonal matrix.

Q: Lemma. Let V be a complex inner product space. Let \(\alpha^* = \alpha^{-1}\) so it is unitary. Then all eigenvalues have unit norm and eigenvectors corresponding to different eigenvalues are orthogonal.
A: Let \(\lambda \in \mathbb{C}, v \in V \setminus \{0\}\) such that \(\alpha(v) = \lambda v\). First \(\lambda \neq 0\) since \(\alpha\) is invertible, and in particular \(\text{Ker}(\alpha) = \{0\}\). Since \(v = \lambda \alpha^{-1}v\), we can compute \[\lambda \langle v,v \rangle = \langle \lambda v,v \rangle = \langle \alpha v,v \rangle = \langle v, \alpha^{-1}v \rangle = \langle v, \frac{1}{\lambda} v \rangle = \frac{1}{\lambda} \langle v , v \rangle\]
So \(|\lambda| = 1\). Further suppose \(\alpha(w) = \mu w, \lambda \neq \mu\). Then \[\lambda \langle v, w \rangle = \langle \lambda v,w \rangle = \langle \alpha v , w \rangle = \langle v, \alpha^{-1} w \rangle = \langle v , \frac{1}{\mu} w \rangle = \frac{1}{\mu} \langle v,w \rangle = \mu \langle v , w \rangle\]

Q: Definition. Let V be a finite dimensional inner product space, and \(\alpha\) an endomorphism of V. Let \(\alpha^* \in L(V)\) be the adjoint map. Then the condition \([...]\) is equivalent to the condition \(\alpha^* = \alpha^{-1}\), and such an \(\alpha\) is an isometry (for \(\mathbb{R}\) it is called orthogonal, and for \(\mathbb{C}\) it is called unitary).
A: \langle \alpha v, \alpha w \rangle = \langle v , w \rangle
Back Extra: This preserves distances and angles.

Q: Theorem. Simultaneous Diagonalisation of Quadratic Forms. Let V be a finite dimensional vector space over \(\mathbb{R}\). Let \(\phi, \psi : V \times V \to \mathbb{R}\) be symmetric bilinear forms. If \(\phi\) is positive definite then there exists a basis \(\beta = \{v_1 \dots v_n\}\) s.t. \([\phi]_{\beta}, [\psi]_{\beta}\) are diagonal
A: Let \(A = [\phi]_{\beta '}\), \(B = [\psi]_{\beta '}\) where \(\beta '\) is standard basis. We can write \(P^T A P = I\) for some invertible P since A is positive definite. Now consider \(P^T B P\) is a symmetric matrix so there exists an orthogonal matrix Q s.t. \(Q^T(P^T B P) Q = D\) where D is diagonal. Now consider \[Q^T (P^T A P)Q = Q^T Q =I = (PQ)^T A PQ = I\] and \((PQ)^T B (PQ) = D\)