Title: Part IA / Vectors and Matrices
Tags: Linear_Maps Vectors Eigenvalues_And_Eigenvectors Complex_Numbers Quadratic_Forms_and_Conics Matrices_and_Linear_Equations Transformation_Groups

Q: Definition. Complex number
A: A number \(z \in \mathbb{C}\) of the form \(z = a+ib, a,b \in \mathbb{R}, i^2 = -1\)

Q: Theorem. Triangle inequality. \(|z_1 + z_2| \leq |z_1|+|z_2|, ||z_1|-|z_2|| \leq |z_1 - z_2|\)
A: For first inequality use a diagram, shortest path is always a straight line.
For second inequality \(z_1 = z_1' - z_2', z_2 = z_2'\), so the first inequality becomes \(|z_1'| \leq |z_1' - z_2'| + |z_2'| \implies |z_1' - z_2'| \geq |z_1'| - |z_2'|\), since order only matters on the RHS we can swap 1 and 2 take the modulus of it and the inequality still holds.

Q: Definition. Exponential function
A: \(\exp (z) = e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{z^n}{n!}\)

Q: Lemma. \(\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{mn} = \sum_{r=0}^{\infty}\sum_{m=0}^ra_{r-m, m}\)
A: \(\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{mn} = a_{00} + a_{01}+a_{02}+ \cdots + a_{10} + a_{11} + \cdots = (a_{00}) + (a_{10} + a_{01}) + \cdots = \sum_{r=0}^{\infty}\sum_{m=0}^ra_{r-m, m}\)

Q: Theorem. \(\exp (z_1)\exp (z_2) = \exp(z_1 + z_2)\)
A: \(\exp(z_1)\exp(z_2) = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{z_1^m}{m!} \frac{z_2^n}{n!} = \sum_{r=0}^{\infty} \sum_{m=0}^{r} \frac{z_1^{r-m}}{(r-m)!} \frac{z_2^m}{m!} = \sum_{r=0}^{\infty}\frac{1}{r!} \sum_{m=0}^{r} \frac{r!}{(r-m)!m!} z_1^{r-m} z_2^m = \sum_{r=0}^{\infty} \frac{(z_1 + z_2)^r}{r!}\)

Q: Definition. Define for all \(z \in \mathbb{C}, \sin z,\cos z\)
A: \(\sin z = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}z^{2n+1}\)
\(\cos z = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}z^{2n}\)

Q: Theorem. \(e^{iz} = \cos z + \sin z\)
A: Proof using Taylor series.

Q: Defintion. Roots of Unity
A: The nth roots of unity are the roots to the equation \(z^n = 1, n \in \mathbb{N}\)

Q: Proposition. If \(\omega = e^{\frac{2\pi i}{n}}, 1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0\)
A: Consider the equation \(z^n = 1\), the coefficient of \(z^{n-1}\) is the sum of all roots. Since the coefficient of \(z^{n-1}\) is 0 then the sum of all roots is 0.

Q: Definition. Complex Logarithm
A: The complex logarithm \(\omega = \log z\) is a solution to \(e^{\omega} = z\). Writing \(z = re^{i\theta}, \log z = \log r + i\theta\). We select \(-\pi < \theta \leq \pi\).

Q: Definition. Complex Power
A: \(z^{\alpha}\) is defined as \(z^{\alpha} = e^{\alpha \log z}\). This can be multi-valued as \(z^{\alpha} = e^{\alpha \log |z|}e^{i \alpha \theta}e^{2in\pi \alpha}\). Once again we restrict \(-\pi < \theta \leq \pi\)

Q: Theorem. De Moivre's Theorem \[\cos n\theta + i \sin n\theta = (\cos \theta + i \sin \theta)^n\]
A: Prove by induction using trig laws. For negative powers just let \(m = -n\) and reduce it to \((\cos m \theta + i \sin m \theta)^{-1}\)

Q: Theorem. Equation of straight line through \(z_0\) and parallel to \(w\)
A: \[z \bar{w } - \bar{z}w = z_0\bar{w} - \bar{z_0}w\]
Rearrange for \(\lambda\) in \(z = z_0 + \lambda w\), then take the conjugate and since \(\lambda\) is real these two expressions are equivalent

Q: Theorem. The general equation of a circle with centre \(c \in \mathbb{C}\) and radius \(\rho \in \mathbb{R}^+\)
A: \[z\bar{z} - \bar{c}z - c\bar{z} = \rho^2 -c\bar{c}\]
Derive using \(|z-c| = \rho\) and \(|z|^2 = z\bar{z}\)

Q: Definition. (Vector) A vector space over \(\mathbb{R}\) or \(\mathbb{C}\) is a collection of vectors \( v \in V\) together with two operations: addition of two vectors and multiplication of a vector with a scalar
A: Vector addition has to satisfy the following axioms:
i) \( \bf a + b = b + a \text{(commutativity)}\)
ii) \(\bf (a + b) + c = a + (b+ c)\text{(associativity)}\)
iii) There is a vector 0 such that \(\bf a + 0 = a \text{(identity)}\)
iv) For all vectors, there is an inverse vector such that \(\bf a + (-a) = 0\)
Scalar multiplication has to satisfy the following axioms:
i) \(\lambda(\bf{a+b}) = \lambda\bf{a} + \lambda\bf{b}\)
ii)\((\lambda + \mu)\bf{a} = \lambda \bf{a} + \mu \bf{a}\)
iii)\(\lambda(\mu \bf{a}) = (\lambda \mu)\bf{a}\)
iv) \(1\bf{a = a}\)

Q: Definition. Unit vector
A: A unit vector is a vector with length 1. We write a unit vector as \(\hat{\bf{v}}\)

Q: Definition. Scalar/dot product
A: \(\bf a \cdot b = |a|b| \cos \theta\), where \(\theta\) is the angle between a and b
It satisfies the following:
i)\(\bf a \cdot b = b \cdot a\)
ii) \(\bf a \cdot a = |a|^2 \geq 0\)
iii) \(\bf a \cdot a = 0 \text{ iff } a = 0\)
iv)If \(\bf a \cdot b = 0\) and \(\bf a,b \neq 0\) then \(\bf a, b\) are perpendicular

Q: Definition. Inner/scalar product
A: In a real vector space V, an inner product is a map \(V \times V \to \mathbb{R}\) that satisfies the following axioms. It is written as \(\bf{x \cdot y}\) or \(\langle \bf x | y \rangle\).
i)\(\bf x \cdot y = y\cdot x\)(symmetry)
ii)\(\bf x \cdot(\lambda y + \mu z) = \lambda x \cdot y + \mu x \cdot z\)(linearity in 2nd argument)
iii)\(\bf x \cdot x \geq 0\) with equality iff \(\bf x = 0\)(positive definite)

Q: Definition. The norm of a vector, \(|a|\)
A: \(|a| = \sqrt{\bf a \cdot a}\)

Q: Theorem. Cauchy-Schwarz inequality. For all \(\bf x, y \in \mathbb{R}^n\), \[\bf |x \cdot y| \leq |x||y|\]
A: Consider the expression \(|\bf x - \lambda y|^2\), we must have \[|\bf x - \lambda y|^2 \geq 0 \implies (x - \lambda y) \cdot (x - \lambda y) \geq 0 \implies \lambda^2|x^2| - \lambda(2x \cdot y) + |x|^2 \geq 0\]
Treat this as a quadratic in \(\lambda\) and since it is non-negative it cannot have 2 real roots so use the discriminant

Q: Corollary. Triangle inequality(Vector Proof)
A: \(\bf |x + y|^2 = (x + y)\cdot (x + y) = |x|^2 + 2x\cdot y + |y|^2 \leq |x|^2 + 2|x||y| + |y|^2 = (|x| + |y|)^2\)

Q: Definition. Vector\cross product
A: For \(\bf a,b \in \mathbb{R}^3\) define the vector product to be \(\bf a \times b = |a||b| \sin \theta \hat{n}\), where \(\bf \hat{n}\) is a unit vector perpendicular to both \(\bf a, b\) in a right-handed sense.
It satisfies the following properties:
i) \(\bf a \times b = -b \times a\)
ii) \(\bf a \times a = 0\)
iii) \(\bf a \times b = 0 \implies a = \lambda b\)
iv) \(\bf a \times (\lambda b) = \lambda(a \times b)\)
v)\(\bf a \times (b + c) = a \times b + a \times c\)

Q: Definition. Scalar triple product
A: \[\bf [a,b,c] = a \cdot(b \times c)\]

Q: Proposition. If a parallelepiped has sides represented by vectors \(\bf a,b,c\) that form a right-handed system, then the volume of the parallelpiped is given by \(\bf[a,b,c]\)
A: The area of the base of the parallelepiped is given by \(\bf |b||c|\sin \theta = |b \times c|\), thus the volume \(\bf = |b \times c||a| \cos \phi = |a \cdot (b \times c) |\), where \(\phi \) is the angle between the base and the vector perpendicular to it.

Q: Theorem. \(\bf a \times (b + c) = a \times b + a \times c\)
A: Let \(\bf d = a \times (b + c) - a \times b - a \times c\), then show that \(\bf d \cdot d = 0 \implies d = 0\)

Q: Theorem. The coefficients \(\lambda, \mu \) are unique when \(\bf r = \lambda a + \mu b\)
A: Suppose that \(\bf r = \lambda a + \mu b = \lambda ' a + \mu ' b\). Take the vector product with a on both sides to get \((\mu - \mu ' \bf) a \times b = 0\). Since \(\bf a \times b \neq 0\), then \(\mu = \mu '\). Repeat for \(\lambda\)

Q: Definition. Linearly independent vectors in \(\mathbb{R}^2\)
A: Two vectors are linearly independent if for \(\alpha, \beta \in \mathbb{R}, \alpha \bf a + \beta b = 0 \iff \alpha = \beta = 0\). In \(\mathbb{R}^2\), two vectors are linearly indepedent if \(\bf a \times b \neq 0\)

Q: Definition. Basis of \(\mathbb{R}^2\)
A: A set of vectors is a basis of \(\mathbb{R}^2\) if it spans \(\mathbb{R}^2\) and are linearly independent

Q: Theorem. If \(\bf a,b,c \in \mathbb{R}^3\) are non-coplanar, i.e. \(\bf a \cdot (b \times c) \neq 0\), then they form a basis of \(\mathbb{R}^3\)
A: For any r, write \(\bf r = \lambda a + \mu b + \nu c\). Perform the scalar product with \(\bf b \times c\) on both sides giving \(\bf r \cdot (b \times c) = \lambda a \cdot (b \times c) + \mu b \cdot(b \times c) + \nu c \cdot(b \times c) = \lambda [a,b,c]\). Thus \(\lambda = \bf \frac{[r,b,c]}{[a,b,c]}\). The values of \(\mu, \nu \) can be found similarly. Thus each r can be written as a linear combination of \(\bf a,b,c\). By the formula derived it follows that \(\bf \alpha a + \beta b + \gamma c = 0 \implies \alpha = \beta = \gamma = 0\).

Q: Definition. Linearly independent vectors
A: A set of vectors \({\bf v_1,v_2,v_3 \cdots v_m}\) is linearly independent if \(\sum_{i=1}^m \lambda_i \bf{v}_i = \bf{0} \implies (\forall i)\lambda_i = 0\)

Q: Definition. Spanning set
A: A set of vectors \(\bf {u_1, u_2, u_3 \cdots u_m} \subseteq \mathbb{R}^n\) is a spanning set of \(\mathbb{R}^n\) if \[(\forall \bf{x} \in \mathbb{R}^n)(\exists \lambda_i)\sum_{i=1}^{m} \lambda_i u_i = x\]

Q: Definition. Basis vectors
A: A basis of \(\mathbb{R}^n\) is a linearly independent spanning set

Q: Definition. Orthonormal basis
A: A basis \(\bf {e_i}\) is orthonormal if \(\bf e_i \cdot e_j = 0\) if \(i \neq j\) and \(\bf e_i \cdot e_j = 1\) for all i,j.

Q: Definition. Dimension of vector space
A: The dimension of a vector space is the number of vectors in its basis

Q: Definition. Scalar product
A: The scalar product of \(\bf x,y \in \mathbb{R}^n\) is defined as \(\bf x \cdot y= \sum x_i y_i\)

Q: Definition. \(\mathbb{C}^n\). It has the same standard basis as \(\mathbb{R}^n\) but the scalar product is defined as \(\bf u \cdot v = \sum u_i^* v_i\).It has the following properties.
A: i) \(\bf u \cdot v = (v \cdot u)^*\)
ii) \(\bf u \cdot(\lambda v + \mu w) = \lambda(u \cdot v) + \mu(u \cdot w)\)
iii) \(\bf u \cdot u \geq 0\) and \(u \cdot u = 0 \iff u = 0\)
iv) \((\lambda \bf u + \mu v) \cdot w = \lambda^* u \cdot w + \mu^* v \cdot w\)

Q: Definition. Vector subspace
A: A vector subspace of a vector space V is a subset of V that is also a vector space under the same operations.
Criterion for \(U \subseteq V\) is a subspace iff:
i) \(\bf x,y \in U \implies (x + y) \in U\)
ii) \(\bf x \in U \implies \lambda x \in U\) for all scalars \(\lambda\)
iii)\(\bf 0 \in U\)
or more concisely:
U is non-empty and \(\forall \bf x,y \in U, (\lambda x + \mu y) \in U\)

Q: Notation. Einstein's summation convention
A: Consider a sum \(\bf x \cdot y = \sum x_i y_i\). The summation convention says we can drop the \(\sum\) and simply write \(\bf x \cdot y = x_i y_i\). The rules are:
i) Suffix appears once in a term: free suffix
ii)Suffix appears twice in a term: dummy suffix and is summed over
iii)Suffix appears three times or more: WRONG

Q: Definition. Kronecker Delta
A: \[\delta_{ij} = \left\{ \begin{array}{11} 1 \; i = j \\ 0 \; i \neq j \end{array} \right .\]\[\left ( \begin{array}{111} \delta_{11} & \delta_{12} & \delta_{13} \\ \delta_{21} & \delta_{22} & \delta_{23} \\ \delta_{31} & \delta_{32} & \delta_{33}\end{array} \right ) = \left( \begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right ) = \bf I\]

Q: Definition. Alternating symbol \(\varepsilon_{ijk}\)
A: Define \[\varepsilon_{ijk} = \left\{ \begin{array}{11} +1 & ijk \text{ is even permutation} \\ -1 & ijk \text{ is odd permutation} \\ 0 & \text{otherwise}\end{array} \right .\]

Q: Proposition. \((\bf a \times b)_i = \varepsilon_{ijk}a_jb_k\)
A: Expansion of formula

Q: Theorem. \(\varepsilon_{ijk} \varepsilon_{ipq} = \delta_{jp} \delta_{kq} - \delta_{jq} \delta_{kp}\)
A: Proof by exhaustion: \[\text{RHS} = \left \{ \begin{array}{11} +1 & \text{if j =p and k = q} \\ -1 & \text{if j=q and k=p} \\ o & \text{otherwise} \end{array} \right .\]
LHS: Summing over i, the only non-zero terms are when \(j,k \neq i\) and \(p,q \neq i\). If j = p and k =q, LHS is \((-1)^2\) or \((+1)^2 = 1\). If j=q and k=p, LHS is \((+1)(-1)\) or \((-1)(+1) = -1\). All other possibilities result in 0.

Q: Proposition. \[\bf a \cdot(b \times c) = b \cdot (c \times a)\]
A: In suffix notation we have \[\mathbf{a \cdot (b \times c) = a_i(b \times c)_i} = \varepsilon_{ijk} b_j c_k a_i = \varepsilon_{jki}b_jc_ka_i = \mathbf{b \cdot (c \times a)}\]

Q: Theorem. Vector triple product \[\mathbf{a \times (b \times c) = (a \cdot c)b - (a \cdot b)c}\]
A: \[\begin{equation} \begin{split} [\mathbf{a \times(b \times c)}]_i = \varepsilon_{ijk}a_j(b \times c)_k \\ = \varepsilon_{ijk}\varepsilon_{kpq}a_jb_pc_q \\ = \varepsilon_{ijk} \varepsilon_{pqk}a_jb_pc_q \\ = (\delta_{ip}\delta_{jq} - \delta_{iq}\delta_{jp})a_jb_pc_q \\ = a_jb_ic_j - a_jc_ib_j \\ = (a \cdot c)b_i - (a \cdot b)c_i\end{split}\end{equation}\]

Q: Theorem. Equation of a straight line through a and parallel to t
A: \[\bf (x-a)\times t = 0\]

Q: Theorem. Equation of a plane through b with normal n, or equation of a plane containing a,b,c
A: \[\bf x \cdot n = b \cdot n\]
\[\mathbf{(x-a)\cdot[(b-a)\times(c-a)]} = 0\]

Q: Proposition. Shortest distance between 2 lines
A: \[s = \left | (\bf a_1 - a_2) \cdot \frac{t_1 \times t_2}{|t_1 \times t_2|}\right |\]

Q: Example. Rotation in \(\mathbb{R}^3\) around the z axis by \(\theta\).
A: Call the rotation R and let \(x' = R(x)\), write \(x = (x,y,z) = (r \cos \phi, r \sin \theta, z)\). Then after the rotation we have, \(x' = (r \cos (\phi + \theta), r \sin (\phi + \theta), z) = (x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta, z)\)
So we represent this with a matrix R such that \(x'_i = R_{ij}x_j\) \[R = \left (\matrix{\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1} \right )\]

Q: Example. Reflection in \(\mathbb{R}^3\) through a plane through the origin with normal \(\bf \hat{n}\)
A: Let the point we are reflecting be \(\bf x\).
The projection of \(\bf x\) onto \(\bf \hat{n}\) is given by \((\bf x \cdot \hat{n})\hat{n}\). So we get \(\bf x' = x - 2(x \cdot \hat{n})\hat{n}\).
In suffix notation we have \(x'_i = x_i - 2x_jn_jn_i\). So the reflection matrix is \(R_{ij} = \delta_{ij} - 2n_in_j\)

Q: Definition. Domain, codomain and image of map
A: Consider sets A and B and mapping \( T : A \to B\) such that each \(x \in A\) is mapped into a unique \(x' = T(x) \in B\). A is the domain of T, B is the co-domain of T.

Q: Definition. Linear map
A: \(T : V \to W\) is a linear map if:
i) \(T(\mathbf{a + b}) = T(\mathbf{a}) + T(\mathbf{b}), \forall \mathbf{a,b} \in V\)
ii) \(T(\lambda \mathbf{a}) = \lambda T(\mathbf{a}), \forall \lambda \in \mathbb{R}\)

Q: Definition. Image and kernel of map
A: The image of a map \(f : U \to V\) is the subset of \(V \{ f(\mathbf{u}) : \mathbf{u} \in U\}\)
The kernel is the subset of \(U \{\mathbf{u} \in U: f(\mathbf{u}) = \mathbf{0}\}\)

Q: Theorem. Consider a linear map \(f : U \to V\), where \(U, V\) are vector spaces. Then im(f) is a subspace of V, and ker(f) is a subspace of U
A: Both are non empty since \(f(\bf{0}) = 0\).
If \(\mathbf{x,y} \in \text{im }(f), \text{ then } \exists \mathbf{a, b} \in U\) such that \(\mathbf{x} = f(\mathbf{a}), \mathbf{y} = f(\mathbf{b})\). Then \(\lambda x + \mu y = \lambda f(\mathbf{a}) + \mu f(\mathbf{b}) = f(\lambda \mathbf{a} + \mu \mathbf{b})\). Now \(\lambda \mathbf{a} = \mu \mathbf{b} \ in U\) since U is a vector space, so there is an element in U that maps to \(\lambda x + \mu y\), so it is in the image and the image is a subspace of V.
Similarly for the kernel except \(f(\lambda \mathbf{x} + \mu \mathbf{y}) = \lambda f(\mathbf{x}) + \mu f(\mathbf{y}) = \lambda \mathbf{0} + \mu \mathbf{0} = \mathbf{0}\). So it is in the kernel and it is a subspace.

Q: Definition. Rank of linear map
A: The rank is the dimension of the image of the map. Denoted \(r(f)\)

Q: Definition. Nullity of linear map
A: The nullity of f, denoted \(n(f)\) is the dimension of the kernel of f.

Q: Theorem. Rank-nullity theorem
A: For a linear map \(f : U \to V\), \[r(f) + n(f) = \dim (U)\]

Q: Definition. Matrix. Let \(\alpha : \mathbb{R}^n \to \mathbb{R}^m\) be a linear map, and \(x' = \alpha(x)\)
A: We get \(x_i' = [\alpha (\mathbf{e}_j)]_i x_j\). We now define \(A_{ij} = [\alpha(\mathbf{e}_j)]_i\), then \(x_i' = A_{ij}x_j\). \(A_{ij}\) is the entry in the ith row of the jth column. We saw that A is an m x n matrix.

Q: Definition. Matrix multiplication
A: Consider maps \(\alpha : \mathbb{R}^l \to \mathbb{R}^n\) and \(\beta : \mathbb{R}^n \to \mathbb{R}^m\). The composition is \(\beta \alpha : \mathbb{R}^l \to \mathbb{R}^m\). Take \(\mathbf{x} \in \mathbb{R}^l \mapsto \mathbf{x''} \in \mathbb{R}^m\). Then \(\mathbf{x''} = (BA) \mathbf{x} = B \mathbf{x'}\), where \(\mathbf{x} = A \mathbf{x}\). Using suffix notation, we have \(x_i'' = (B \mathbf{x'})_i = b_{ik}x_k' = B_{ik}A_{kj}x_j.\) But \(x_i'' = (BA)_{ij}x_j\). So \[(BA)_{ij} = B_{ik}A_{kj}\]
Generally an m x n matrix multiplied by an n x l matrix gives an m x l matrix \((BA)_{ij}\) is given by the ith row of B dotted with the jth column of A

Q: Definition. Transpose of matrix
A: If A is an m x n matrix, the transpose \(A^T\) is an n x m matrix defined by \((A^T)_{ij} = A_{ji}\)

Q: Proposition. \((AB)^T = B^TA^T\)
A: \((AB)^T_{ij} = (AB)_{ji} = A_{jk}B_{ki} = B_{ki}A_{jk} = (B^T)_{ik}(A^T)_{kj} = (B^TA^T)_{ij}\)

Q: Definition. Hermitian conjugate and Hermitian matrix
A: \[A^{\dagger} = (A^T)^*\]
A matrix is Hermitian if \[A^{\dagger} = A\] so the diagonal is real

Q: Definition. Symmetric matrix
A: \[A^T =A\]

Q: Defintion. Anti symmetric matrix
A: \[A^T = -A\]
The diagonals are all 0

Q: Defintiion. Skew-Hermitian Matrix
A: \[A^{\dagger} = -A\]The diagonals are pure imaginary

Q: Definition. Trace of matrix
A: \[\text{tr} (A) = A_{ii}\]
The sum of the diagonal

Q: Proposition. tr(BC) = tr(CB)
A: \(\text{tr}(BC) = BC_{ii} = B_{ik}C_{ki} = C_{ki}B_{ik} = (CB)_{kk} = \text{tr}(CB)\)

Q: Proposition. Decomposition of a square matrix
A: 1. Split into symmetric and antisymmetric parts \[B_{ij} = \frac{1}{2}(B_{ij} + B_{ji}) + \frac{1}{2}(B_{ij} - B_{ji})\]
2. Split the symmetric part into an isotropic part (a scalar multiple of the identity) and a trace-less part
\[S_{ij} = \frac{1}{n}\text{tr}(S) \delta_{ij} + (S_{ij} - \frac{1}{n} \text{tr}(S)\delta_{ij})\]3.Combining these
\[B = \frac{1}{n} \text{tr}(B) I + \left \{ \frac{1}{2} (B + B^T) - \frac{1}{n}\text{tr}(B)I\right \} + \frac{1}{2}(B - B^T)\]

Q: Definition. Inverse of matrix
A: If \(BA = I\) then we say B is a left inverse and if \(AC = I\) then we say C is a right inverse . If A is square then \[B = B(AC) = (BA)C = C\]
So we have for a square matrix \[AA^{-1} = A^{-1}A = I\]

Q: Definition. Orthogonal and unitary matrices
A: A real n x n matrix is orthogonal if \(A^T A = AA^T = I\) or \(A^T = A^{-1}\)
A complex n x n matrix is unitary if \(U^{\dagger} U = U U^{\dagger} = I\) or \(U^{\dagger} = U^{-1}\)
Note that an orthogonal matrix A satisfies \(A_{ik}(A_{kj}^T) = \delta_{ij} \) or \(A_{ik}A_{jk} = \delta_{ij}\)
This is saying that the scalar product of two distinct rows is 0 and the scalar product of a row with itself is 1. So the rows - and columns by considering \(A^T\) - of an orthogonal matrix form an orthonormal set

Q: Definition. Levi-Civita symbol
A: \[\varepsilon_{j_1j_2j_3 \cdots j_n} = \left \{ \begin{matrix} +1 & \text{if } j_1j_2\cdots j_n \text{ is an even permutation} \\ -1 & \text{if it is an odd permutation} \\ 0 & \text{if any 2 of them are equal} \end{matrix}\right .\]

Q: Definition. Determinant
A: \[\det A = \sum_{\sigma \in S_n} \varepsilon(\sigma)A_{\sigma(1)1} A_{\sigma(2)2} \dots A_{\sigma(n)n} = \varepsilon_{j_1j_2\cdots j_n}A_{j_11} A_{j_22} \cdots A_{j_nn}\]

Q: Proposition. \(\det A = \det A^T\)
A: Take a single term \(A_{\sigma(1)1}A_{\sigma(2)2}\dots A_{\sigma(n)n}\) and let \(\rho \in S_n\) be another permutation, then \[A_{\sigma(1)1}A_{\sigma(2)2}\dots A_{\sigma(n)n} = A_{\sigma(\rho(1))\rho(1)}A_{\sigma(\rho(2))\rho(2)}\dots A_{\sigma(\rho(n))\rho(n)}\]
as this is just a reordering. Now let \(\rho = \sigma^{-1}\) and note that \(\epsilon(\sigma) = \epsilon(\rho)\). Then \[\det A = \sum_{\rho \in S_n} \epsilon(\rho)A_{1 \rho(1)}A_{2\rho(2)}\dots A_{n\rho(n)} = \det A^T\]

Q: Proposition. If matrix B is formed by multiplying every element in a single of A by a scalar \(\lambda\) then \(\det B = \lambda \det A\), \(\det \lambda A = \lambda^n \det A\)
A: Each term in the sum is multiplied by \(\lambda\) so the whole sum is multiplied by \(\lambda^n\)

Q: Proposition. If 2 rows of A are identical, the determinant is 0
A: WLOG suppose columns 1 and 2 are the same. Then \[\det A = \sum_{\sigma \in S_n} \varepsilon(\sigma)A_{\sigma(1)1}A_{\sigma(2)2}\dots A_{\sigma(n)n}\]
Now write an arbitrary \(\sigma\) in the form \(\sigma = \rho(1 \; 2)\). Then \(\varepsilon(\sigma) = \varepsilon(\rho)\varepsilon((1 \; 2)) = - \varepsilon(\rho)\), so \[\det A = \sum_{\rho \in S_n} - \varepsilon(\rho)A_{\rho(2)1}A_{\rho(1)2}\dots A_{\rho(n)n}\]
But since columns 1 and 2 are identical then \(A_{\rho(2)1} = A_{\rho(2)2}\) and same for \(A_{\rho(1)2} = A_{\rho(1)1}\). So \(\det A = - \det A \implies \det A = 0\)

Q: Proposition. If 2 rows or columns of a matrix are linearly dependent, then the determinant is 0.
A: Suppose in A, column(r) + \(\lambda\)column(s) = 0. Define \[B_{ij} = \left \{ \begin{matrix}A_{ij} & j \neq r \\ A_{ij} + \lambda A_{is} & j=r \end{matrix} \right .\]
Then \(\det B = \det A + \lambda \det (\text{matrix with column r = column s}) = \det A\). Then we can see that the rth column of B is all zeroes. Ao each term in the sum contains one zero and \(\det A = \det B = 0\).
Linear dependence is required to show that the determinants are 0, otherwise we just get that \(\det A = \det B\) so we can add linear multiples of columns to other columns without changing the determinant.

Q: Corollary. Swapping two rows or columns of a matrix negates the determinant.
A: We do the column case only. Let \(A = (\mathbf{a_1 \dots a_i \dots a_j \dots a_n})\). Then \[\det(\mathbf{a_1 \dots a_i \dots a_j \dots a_n}) = \det(\mathbf{a_1 \dots a_i + a_j \dots a_j \dots a_n}) = \det(\mathbf{a_1 \dots a_i + a_j \dots a_j - (a_i + a_j) \dots a_n}) = \det(\mathbf{a_1 \dots a_i + a_j \dots -a_i \dots a_n}) = \det(\mathbf{a_1 \dots a_j \dots -a_i \dots a_n}) = -\det(\mathbf{a_1 \dots a_j \dots a_i \dots a_n})\]
Or we can use the fact that the sign of a transposition is -1 and that the sign is multiplicative.

Q: Proposition. \(\det AB = \det A \det B\)
A: First note that \(\sum_{\sigma} \varepsilon(\sigma) A_{\sigma(1)\rho(1)}A_{\sigma(2)\rho(2)} = \varepsilon(\rho) \det A\), so swapping columns or rows an even/odd number of times gives a factor \(\pm 1\) respectively. We now write \(\sigma = \mu \rho\). Now \[\det AB = \sum_{\sigma} \varepsilon(\rho)(AB)_{\sigma(1)1} (AB)_{\sigma(2)2} \dots (AB)_{\sigma(n)n} = \sum_{\sigma} \varepsilon(\sigma) \sum_{k_1, k_2, \dots k_n}^n A_{\sigma(1)k_1}B_{k_11} \dots A_{\sigma(n)k_n} B_{k_nn} = \sum_{k_1, \dots , k_n} B_{k_11} \dots B_{k_nn} \sum_{\sigma} \varepsilon(\sigma) A_{\sigma(1)k_1} A_{\sigma(2)k_2} \dots A_{\sigma(n)k_n}\]
We only care if the \(k_i\) are distinct as otherwise the second sum will be 0 as two columns will be the same, so we only have to conisder when \(k_i = \rho(i)\). Then we can write \[\det AB = \sum_{\rho} B_{\rho(1)1} \dots B_{\rho(n)n} \sum_{\sigma} \varepsilon(\sigma)A_{\sigma(1)\rho(1)} \dots A_{\sigma(n)\rho(n)} = \sum_{\rho} B_{\rho(1)1} \dots B_{\rho(n)n}(\varepsilon(\rho)\det A)= \det A \sum_{\rho} \varepsilon(\rho)B_{\rho(1)1 }\dots B_{\rho(n)n} = \det A \det B\]

Q: Theorem. Laplace expansion formula
A: \[\det A = \sum_{j=1}^n A_{ji}\Delta_{ji}\]

Q: Definition. Minor and cofactor of an n x n matrix A
A: Define \(A^{ij}\) to be the \((n-1) \times (n-1)\) matrix in which row i and column j of A have been removed.
The minor of the ijth element of A is \(M_{ij} = \det A^{ij}\)
The cofactor of the ijth element of A is \(\Delta_{ij} = (-1)^{i+j}M_{ij}\)

Q: Corollary. If A is orthogonal, \(\det A = \pm 1\)
A: \[AA^T = I\]
\[\det AA^T = \det I = \det A \det A^T = (\det A)^2 = 1\]

Q: Lemma. \(\sum A_{ik}\Delta_{jk} = \delta_{ij} \det A\)
A: If \(i \neq j\), consider an n x n matrix B, which is identical to A except the jth row is replaced by the ith row of A. So \(\Delta_{jk}\) of B = \(\Delta_{jk}\) of A since \(\Delta_{jk}\) does not depend on the elements in row j. Since B has a duplicate row we know that \[0 = \det B = \sum_{k=1}^n B_{jk}\Delta_{jk} = \sum_{k=1}^n A_{ik}\Delta_{jk}\]
If i = j, then the expression is det A by the Laplace expansion formula

Q: Theorem. If \(\det A \neq 0\), then \(A^{-1}\) exists and is given by \[(A^{-1})_{ij} = \frac{\Delta_{ji}}{\det A}\]
A: \((A^{-1})_{ik} A_{kj} = \frac{\Delta_{ik}}{\det A}A_{kj} = \frac{\delta_{ij} \det A}{\det A} = \delta_{ij}\)
So \(A^{-1}A = I\)

Q: Definition. Homogeneous equation
A: If \(Ax = b\) where A is an n x n matrix, x and b are n x 1 column vectors
If b = 0, then the system is homogeneous
Otherwise its inhomogeneous

Q: Definition. Column and row rank of linear map
A: Column rank of a matrix is the maximum number of linearly independent columns.
Row rank is the maximum number of linearly independent rows.

Q: Theorem. The column rank and row rank are equal for any m x n matrix
A: Let r be the row rank of A. Write the biggest set of linearly independent rows as \(\mathbf{v_1^T, v_2^T, \dots v_r^T}\) or in component form \(\mathbf{v_k^T} = (v_{k1}, v_{k2}, \dots u_{kn})\) for \(k = 1,2,\dots r\)
Now denote the ith row of A as \(\mathbf{r_i^T} = (A_{i1}, A_{i2}, \dots A_{in})\).
Now we can write every of A as a linear combinations of the \(\mathbf{v's}\) as they are the maximum number of independent rows.
Write \[\mathbf{r_i^T} = \sum_{k=1}^r C_{ik} \mathbf{v_k^T}\] for \(C_{ik}, 1 \leq i \leq m, 1 \leq k \leq r\)
Now the elements of A are \[A_{ij} = (\mathbf{r_i})^T_j = \sum_{k=1}^r C_{ik}(\mathbf{v_k})_j\]
So every column of A can be written as a linear combination of the r column vectors \(\mathbf{c_k}\). Then the column rank of A \(\leq r\), the row rank of A.
Apply the same argument to \(A^T\) to see that they are equal.

Q: Definition. Multiplicity of root
A: The root \(z = w\) has multiplicity k if \((z-w)^k\) is a factor of \(p(z)\) but \((z-w)^{k+1}\) is not

Q: Definition. Eigenvector and Eigenvalue
A: Let \(\alpha : \mathbb{C}^n \to \mathbb{C}^n\) be a linear map with associated matrix A. Then \(\bf x \neq 0\)is an eigenvector of A if \(A\mathbf{x} = \lambda \mathbf{x}\).
\(\lambda\) is the associated eigenvalue.

Q: Theorem. \(\lambda\) is an eigenvalue of A iff \[\det (A - \lambda I) = 0\]
A: Suppose that \(\lambda\) is an eigenvalue and \(\mathbf{x}\) is the associated eigenvector. Rearrange to \((A - \lambda I)\mathbf{x = 0}\) and thus \( \mathbf{x} \in \text{ker }(A - \lambda I)\). But \(\mathbf x \neq 0\). So the kernel is non-trivial and \(\det (A - \lambda I) = 0\) as otherwise there are inverses and kernel is trivial. Otherdirection is similar.

Q: Definition. Characteristic equation of matrix
A: \[\det (A - \lambda I) = 0\]

Q: Definition. Characteristic polynomial of matrix
A: The characteristic polynomial of A is \[p_A (\lambda) = \det (A - \lambda I) = \varepsilon_{j_1j_2\dots j_n} (A_{j_1 1} - \lambda \delta_{j_11}) \cdots (A_{j_n}n - \lambda \delta_{j_n n}) = c_0 + c_1 \lambda + \cdots + c_n \lambda^n\]i) the polynomial has degree n so an n x n matrix has n eigenvalues
ii) if A is real then so are the coefficients of the polynomial, so eigenvalues are either real or come in complex conjugate pairs
iii) \(c_n = (-1)^n\) and \(c_{n-1} = (-1)^{n-1}(A_{11} + A_{22} + \cdots + A_{nn}) = (-1)^{n-1} \text{tr} (A) \). But it is also the sum of roots so \[\text{tr}(A) = \lambda_1 + \lambda_2 + \cdots + \lambda_n\]
Finally, \(c_0 = p_A(0) = \det (A)\). Also \(c_0\) is the product of all roots, i.e. \(c_0 = \lambda_1 \lambda_2 \dots \lambda_n\). So \[\det A = \lambda_1 \lambda_2 \dots \lambda_n\]

Q: Definition. Eigenspace
A: \(E_{\lambda}\) is the kernel of the matrix \(A - \lambda I\), i.e. the set of eigenvectors with eigenvalue \(\lambda\)

Q: Definition. Algebraic multiplicity of eigenvalue.
A: \(M(\lambda)\) or \(M_{\lambda}\) of an eigenvalue \(\lambda\) is the multiplicity of \(\lambda\) in \(p_A (\lambda) = 0\). By the fundamental theorem of algebra \[\sum_{\lambda} M(\lambda) = n\]If \(M(\lambda) > 1\), then the eigenvalue is degenerate

Q: Definition. Geometric multiplicity of eigenvalue
A: \(m(\lambda)\) of an eigenvalue is the dimension of the eigenspace, i.e. the maximum number of linearly independent eigenvectors with eigenvalue \(\lambda\).

Q: Definition. Defect of eigenvalue
A: \[\Delta_{\lambda} = M(\lambda) - m(\lambda)\]

Q: Theorem. Suppose n x n matrix A has distinct eigenvalues \(\lambda_1, \lambda_2 \dots , \lambda_n\). Then the corresponding eigenvectors \(\mathbf(x_1, x_2, \dots ,x_n)\) are linearly independent
A: Suppose they are linearly dependent. So we can find \(d_i\) such that \[d_1 \mathbf{x_1} + \dots + d_r \mathbf{x_r} = 0\]Suppose that this is the shortest non-trivial linear combination that gives 0. Now apply \((A - \lambda I)\) to the whole equation to obtain \[d_1 (\lambda_1 - \lambda_1)\mathbf{x_1} + \cdots + d_r(\lambda_r - \lambda_1)\mathbf{x_r} = 0\]. Since the first term is 0 we have found a shorter linear combination that gives 0 there is a contradiction.

Q: Theorem. Consider a linear map associated with a matrix A, and P a matrix associated with a change of basis\[\tilde{A} = P^{-1}AP\]
A: \[\begin{matrix} \mathbf{u'} = A\mathbf{u} \\ P \mathbf{\tilde{u}'} = AP \mathbf{\tilde{u}} \\ \mathbf{\tilde{u}'} = P^{-1}AP\mathbf{u} = \tilde{A}\mathbf{u}\end{matrix}\]

Q: Theorem. Denote vector as \(\bf u\) with respect to \(\{ \mathbf{e_i}\}\) and \(\bf \tilde{u}\) with respect to \(\{\mathbf{\tilde{e}_i}\}\). Then \[\mathbf{u} = P \mathbf{\tilde{u}} \; \; \; \; \tilde{\mathbf{u}} = P^{-1}\mathbf{u}\]
A: \(\mathbf{u} = \sum_{i=1}^n \tilde{u}_i \tilde{\mathbf{e}_i} = \sum_{i=1}^n u_i \mathbf{e_i}\)
Using the transformation matrix we get that \[\begin{matrix} \mathbf{u} = \sum_{j=1}^n \tilde{u}_j \sum_{i=1}^n P_{ij} \mathbf{e_i} \\ \;\;\;\;\;\;\;= \sum_{i=1}^n \left ( \sum_{j=1}^n P_{ij}\tilde{u}_j\right ) \mathbf{e_i}\end{matrix}\]By comparison we get that \[u_i = \sum_{j=1}^n P_{ij}\tilde{u}_j\]

Q: Theorem. Given \(\alpha : \mathbb{C}^m \to \mathbb{C}^n, x \mapsto x'\). Suppose \(\mathbb{C}^m\) changes basis from \(\{\mathbf{e_i}\} \to \{\mathbf{\tilde{e_i}}\}\) and \(\mathbb{C}^n\) from \(\bf \{f_i\} \to \{\tilde{f_i}\}\). Such that \(\bf x = P\tilde{x}, x' = R\tilde{x}'\)where P is an m x m matrix and R an n x n matrix. Then \[\tilde{A} = R^{-1}AP\]
A: \[R\mathbf{\tilde{x}}' = AP\mathbf{\tilde{x}}\]\[\mathbf{\tilde{x}}' = R^{-1}AP\mathbf{\tilde{x}}\]

Q: Definition. Similar matrices
A: Two matrices are similar if there exists an invertible matrix P such that \[B = P^{-1}AP\]i.e. they represent the same map under different bases

Q: Definition. Diagonalizable matrices
A: If it is similar to a diagonal matrix.
Equivalent to saying the eigenvectors form a basis of \(\mathbb{C}^n\).

Q: Theorem. Let \(\lambda_1 , \lambda_2, \dots \lambda_r\) with \(r \leq n\) be the distinct eigenvalues of A. Let \(B_1, B_2, \dots B_r\) be the bases of the corresponding eigenspaces. Then the set \(B = \bigcup_{i=1}^r B_i\) is linearly independent
A: Write \(B_1 = \{x_1^{(1)}, x_2^{(1)}, \dots x_{m(\lambda_1)}^{(1)}\}\) where \(m(\lambda_1) = \text{dim}(E_{\lambda_1})\).
Consider the equation
\[\sum_{i=1}^r \sum_{j=1}^{m(\lambda_i)} \alpha_{ij}x_j^{(i)} = 0\]First sum goes over all eigenspaces and the second goes over the basis vectors in the eigenspace. Now apply the matrix
\[\prod_{k=1,2,\dots,\overline{K}, \dots r}(A-\lambda_kI)\]Where we've missed out some K. This means that for all the terms except the \(x_j^{(k)}\) terms we will get 0. So the sum becomes \[\sum_{j=1}^{m(\lambda_K)} \alpha_{Kj}\left [ \prod_{k=1,2,\dots, \overline{K}, \dots ,r} (\lambda_K - \lambda_k)\right ]x_j^{(k)} = 0\]So since the \(x_j^{(k)}\)are part of a basis, they are linearly independent so all the coefficents are 0. But we picked K arbitrarily so all the coefficients in the first equation are 0 so B is linearly independent.

Q: Theorem. Any \(2 \times 2\) complex matrix A is similar to exactly one of \[\left (\begin{matrix} \lambda_1 & 0 \\ 0 & \lambda_2\end{matrix} \right ), \left (\begin{matrix} \lambda & 0 \\ 0 & \lambda\end{matrix} \right ), \left (\begin{matrix} \lambda & 1 \\ 0 & \lambda\end{matrix} \right )\]
A: i) If 2 distinct eigenvalues then linearly independent eigenvectors so use P formed from eigenvectors as its columns.
ii) If 1 eigenvalue but eigenspace has dimension 2 then use the eigenvectors as the new basis.
iii) If 1 eigenvalue and eigenspace only has dimension 1, then create a basis using the eigenvector and transform to get \[ \left ( \begin{matrix} \lambda & \alpha \\ 0 & \beta \end{matrix} \right )\]Then show that \(\bf u = (\tilde{A} - \lambda I )w\) is an eigenvector of \(\tilde{A}\) with eigenvalue \(\lambda\). Then change basis to u and w.
So P sends basis to \(\{ \bf v, w\}\), and then Q sends to basis \(\{ \bf u,w \}\). So the similarity transofmration is \((PQ)^{-1} A (PQ)\)

Q: Proposition. The canonical form, or Jordan normal form exists for any \(n \times n \) matrix A. Specifically there exists a similarity transofmr such that A is similar to \(\tilde{A}\) that satisfies the following properties:
A: i) \(\tilde{A}_{ii} = \lambda_i\)
ii) \(\tilde{A}_{i, i+1} = 0 \text{ or } 1\)
iii)\(\tilde{A}_{ij} = 0\) otherwise

Q: Theorem. Cayley-Hamilton. Every square complex matrix satisfies its own characteristic equation.
A: Characteristic polynomial is \(f(X) = (a_1 - X)(a_2 - X) \cdots (a_n - X)\).
Then let \(A = diag(a_1,a_2,\dots, a_n)\). Then we see that \(f(A) = (a_1 I - A)(a_2 I -A) \dots (a_nI - A) = diag(0,a_2,\dots,a_n)diag(a_1, 0, a_2, \dots ,a_n) \dots diag(a_1, \dots, a_{n-1}, 0) = diag(0,0,\dots0) = 0\)
Then it must be true for any matrix similar to a diagonal matrix. To see this note that if \(G = PAP^{-1}\), then \[(G-\lambda I) = (PAP^{-1} - \lambda I) = P(A-\lambda I)P^{-1}\]We know that G and A have the same characteristic polynomials, say \(f(\lambda)\). Then \[f(G) = f(PAP^{-1}) = Pf(A)P^{-1} = P0P^{-1} = 0\]
For \(2 \times 2\) matrices similar to \(B = \left ( \begin{matrix} \lambda & 1 \\ 0 & \lambda\end{matrix}\right )\). We see that \(p_B(z) = \det (B-zI) = (\lambda - z)^2\). Then \(p_B{B} = (\lambda I - B)^2 = \left( \begin{matrix} 0 & -1 \\ 0 & 0 \end{matrix}\right )^2 = \bf 0\).

Q: Theorem. The eigenvalues of a Hermitian matrix H are real
A: Start with \(H\mathbf{v} = \lambda \mathbf{v}\), then pre-multiply by \(\bf v^{\dagger}\) and take Hermitian conjugate of both sides. Show that \(\lambda = \lambda^*\).

Q: Theorem. The eigenvectors of a Hermitian matrix H corresponsing to distinct eigenvalues are orthogonal
A: Let v,w be eigenvectors with eigenvalues \(\lambda, \mu\) .
\(v^{\dagger}(Aw) = (Av)^{\dagger}w\)
\(v^{\dagger}(\mu w) = (\lambda v)^{\dagger} w \)
\(\mu v^{\dagger} w = \lambda v^{\dagger} w\)
since \(\lambda \neq \mu, v^{\dagger}w = 0\) so the eigenvectors are orthogonal

Q: Proposition. Gram-Schmidt Orthogonalization. Supposewe have a set \(B = \{w_1, w_2 ,\dots , w_r\}\)of linearly independent vectors. We want to find an orthogonal set \(\tilde{B} = \{v_1, v_2, \dots , v_r\}\)
A: Define the projection of w onto v by \(\mathcal{P}_v(w) = \frac{\langle v | w \rangle}{\langle v | v \rangle} v\)
i) \(v_1 = w_1\)
ii) \(v_2 = w_2 - \mathcal{P}_{v_1}(w_2)\)
iii) \(v_3 = w_3 - \mathcal{P}_{v_1}(w_3) - \mathcal{P}_{v_2}(w_3)\)
and so on....
At each step we substract out the components of \(v_i\) that belong to the space of \(\{ v_1, \dots, v_{k-1}\}\), this ensures that all the vectors are orthogonal.

Q: Definition. Normal matrix
A: Commutes with its own Hermitian conjugate. i.e. \[N N^{\dagger} = N^{\dagger}N\]

Q: Definition. Sesquilinear, Hermitian and quadratic forms
A: Sesquilinear form: \(F = \mathbf{x}^{\dagger}A\mathbf{x} = x_i^*A_{ij}x_j\)
If A is Hermitian, this is a Hermitian form.
If A is real symmetrix, this is a quadratic form.

Q: Theorem. Hermitian forms are real
A: \((\mathbf{x}^{\dagger}H\mathbf{x})^* = (\mathbf{x}^{\dagger}H\mathbf{x})^{\dagger} = \mathbf{x}^{\dagger}H^{\dagger}\mathbf{x} = \mathbf{x}^{\dagger}H\mathbf{x}.\) So it is equal to its complex conjugate and is therefore real.

Q: Definition. Quadric
A: An n-dimensonal surface defined by the zero of a real quadratic polynomial, i.e.
\[\mathbf{x}^TA\mathbf{x} + \mathbf{b}^T\mathbf{x} + c = 0\]Where A is a real n x n matrix, \(\bf x,b\) are n-dimensional column vectors and c is a constant scalar.

Q: Proposition. Conic sections(n=2). Which curves are obtained from different cases of eigenvalues?
A: We can obtain \(\lambda_1x_1^2 + \lambda_2x_2^2 = k\)
i) \(\lambda_1\lambda_2 > 0\): ellipses with axes coinciding with eigenvectors of S.
ii) \(\lambda_1\lambda_2 < 0\): say \(\lambda_1 = k/a^2 > 0, \lambda_2 = -k/b^2 <0\). So we obtain the hyperbola;\[\frac{x_1^2}{a^2} - \frac{x_2^2}{b^2} = 1\]iii)\(\lambda_1\lambda_2 = 0:\) say \(\lambda_2 = 0, \lambda_1 \neq 0\). In this case our symmetric matrix S is not invertible and we cannot shift our origin as above. From our initial equation we have; \[\lambda_1(x_1')^2 + b_1'x_1' + b_2'x_2' + c=0\]
Complete the square to remove the \(x_1'\) and constant term to get \(\lambda_1x_1^2 + b_2x_2 = 0\) which is a parabola.
We assumed \(b_2' \neq 0\), if this is 0 we obtain 0,1 or 2 straight lines

Q: Proposition. How to obtain simple version of a Hermitian form
A: We know that any Hermitian matrix can be diagonalized with a unitary transformatoin. So \(F(\mathbf{x}) = \mathbf{x}^{\dagger}H\mathbf{x} = \mathbf{x}^{\dagger}UDU^{\dagger}\mathbf{x}\). Write \(\mathbf{x}' = U^{\dagger}\mathbf{x}\). So \(F = (\mathbf{x}')^{\dagger}D\mathbf{x'}\), where \(D = diag(\lambda_1, \lambda_2,\dots ,\lambda_n)\).
We know that \(\mathbf{x}'\) is the vector \(\mathbf{x}\) relative to the eigenvector basis. So \[F(\mathbf{x}) = \sum_{i=1}^n \lambda_i |x_i'|^2\]The eigenvectors are known as the principal axes.

Q: Proposition. How to transform a quadric to a quadratic form
A: For any A, the anti-symmetric matrix has \(\mathbf{x}^TA\mathbf{x} = 0\) so we can split any matrix into symmetric and anti-symmetric parts and only keep the symmetric part. So we can have \[\mathbf{x}^TS\mathbf{x} + \mathbf{b^T x} +c = 0\]with S symmetric. Now that S is real and symmetric, we can diagonalize it using \(S = QDQ^T\) with D diagonal. We write \(\mathbf{x}' = Q^T\mathbf{x}\) and \(\mathbf{b}' = Q^T \mathbf{b}\). So we have \[(\mathbf{x}')^TD\mathbf{x}' + (\mathbf{b}')^T\mathbf{x}' + c = 0\]If S is invertible, i.e. with no zero eigenvalues, then write \(\mathbf{x}'' = \mathbf{x}' + \frac{1}{2}D^{-1}\mathbf{b}'\) which shifts the origin to eliminate the linear term \((\mathbf{b}')^T\mathbf{x'}\) and finally have \[\mathbf{x}^TD\mathbf{x} = k\]

Q: Definition. Conic section
A: Eccentricity and scale are properties of a conic section that satisfy the following;
Let the foci of a conic section be \((\pm ae, 0)\) and the directrices be \(x = \pm a/e\).
A conic section is the set of points whose distance from focus is \(e \times \text{distance from directrix which is closer to that of focus}\)(except when e = 1, where we take the distance to the other directrix)

Q: Definition. Minkowski inner product
A: \(\langle \bf x | y \rangle\ = x^TJy\) where \(J = \left ( \begin{matrix} 1 & 0 \\ 0 & -1\end{matrix}\right)\). So \(\langle \mathbf{x | y} \rangle = x_1y_1 - x_2y_2\).

Q: Definition. Lorentz matrix. \[B_v = \frac{1}{\sqrt{1 - v^2}} \left (\begin{matrix} 1 & v \\ v & 1\end{matrix} \right)\]Here \(|v| < 1\), where we have chosen units in which the speed of light is equal to 1.
A: We have \(B_v = H_{tanh^{-1}v}\).
\(H_{\alpha} = \left( \begin{matrix} \cosh \alpha & \sinh \alpha \\ \sinh \alpha & \cosh \alpha \end{matrix}\right)\) is a hyperbolic rotation

Q: Definition. Lorentz group
A: The group of all Lorentz matrices under matrix multiplication.
For the closure axiom we have \(B_{v_1} B_{v_2} = B_{v_3}\) where \[v_3 = \tanh(\tanh^{-1}v_1 + \tanh^{-1}v_2) = \frac{v_1 + v_2}{1+v_1v_2}\]This set of all of these matrices are transformations which preserve the Minowski inner product