f : X -> Y g : Y -> X if inverse exists, g = f^{-1} g(f(x)) = x for all x \in X f(g(x)) = y for all y \in Y For simplicity's sake, we will require bijectivity to define the inverse, although degenerate cases (i.e. non-injective) can be defined. Matrix Inverse A := mxn matrix. Ax where a is nx1 matrix. A can be considered as a function from R^n to R^m. Definition: nxn matrix A is invertible iff there exists B nxn such that AB = BA = I_n. A^{-1} := B. Thm: If A, B are inverses s.t. AB = I_n, BA = I_n. A = [ a1 | a2 | ... an ] B = [ b1 | b2 | ... bn ] AB = [Ab1 | Ab2 | ... Abn ] Let e_i = [ 0 0 ... 1 ... 0 ] where 1 is in the ith position. This gives systems Ab1 = e1, Ab2 = e2 ... Each can be solved like a standard augmented matrix. However, we can solve like [A | e1 | e2 | e3 ... ] (*) Two possibilities: - n pivots (every column has pivot) Reduced echelon form is I_n Right matrix = B = A^{-1} - A_j where j > i to solve [ A | I_n ], we get [ U | L^{-1} ] U is invertible <=> all diagonal elements of U are non-zero <=> every column of U has a pivot column L is always invertible, so iff U is invertible, A = LU is invertible. Transpose A := mxn matrix. A^T = B B := nxm where b_ji = a_ij A : R^n -> R^m B : R^m -> R^n (Not inverse properties) If A is invertible, then A^T is invertible, and (A^{-1})^T = (A^T)^{-1} But why? (1) If A, B are invertible, AB is invertible, and: (AB)^{-1} = B^{-1}A^{-1} [why??] [this should verify the previous identity] (2) (AB)^T = B^T A^T [could be proved by brute calculation] Definition: nxn matrix A is symmetric if A = A^T If A is symmetric and invertible, A = LU = LDL^T (Thm!) Then, D would be invertible. If A not invertible, U not invertible, and D doesn't need to be invertible. This is Cholesky decomposition. "Keeps the symmetry" (?) D is a diagonal (and therefore symmetric) matrix. Chapter 2 --------- Vector space is a collection V of objects called vectors, a binary addition operator, and an operator to multiply a vector and a scalar (defined in R or C) (u + v) + w = u + (v + w) a(u + v) = au + av + Some more rules (probably commutative?) (a+b)u = au + bu. Gives existence of 0 vector. +, * must be closed under V. Ex: Let V = polynomials degree <= 2. Ex: Upper-diagonal 2x2 matrices Ex: R^2 Ex: Subspace of R^2 Not ex: Line in R^2 not containing origin.