Andrew Thangaraj
Aug-Nov 2020
\(n\times n\): square matrix
Simplest is identity
Diagonal matrices
As many zeroes as possible off diagonal
\(m\times n\) matrices
\(T:V\to W\), dim \(V=n\), dim \(W=m\)
Basis for \(V\)
Basis of null \(T\): \(\{u_1,\ldots,u_{n-r}\}\), extend to basis of \(V\)
\(B_V=\{v_1,\ldots,v_r,u_1,\ldots,u_{n-r}\}\), \(r\): rank
Basis for \(W\)
Basis of range \(T\): \(\{T(v_1),\ldots,T(v_r)\}\), extend to basis of \(W\)
\(B_W=\{T(v_1),\ldots,T(v_r),w_1,\ldots,w_{m-r}\}\)
Matrix of \(T\) w.r.t. \(B_V\), \(B_W\)
\(\begin{bmatrix} \vdots&\cdots&\vdots&\vdots&\cdots&\vdots\\ T(v_1)&\cdots&T(v_r)&T(u_1)&\cdots&T(u_{n-r})\\ \vdots&\cdots&\vdots&\vdots&\cdots&\vdots \end{bmatrix}=\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}\)
\(A:m\times n\) matrix, rank: \(r\), represents \(T:F^n\to F^m\) in standard basis
There exist elementary row operations \(E_i\), col operations \(F_j\) s.t.
\(\left(\prod_i E_i\right) A \left(\prod_jF_j\right)=\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}\)
\(S_L=\prod_i E_i\): \(m\times m\) invertible
\(S_R=\prod_jF_j\): \(n\times n\) invertible
\(A=S^{-1}_L\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}S^{-1}_R\)
Change of basis
Basis for \(F^n\): columns of \(S_R\)
Basis for \(F^m\): columns of \(S^{-1}_L\)
\(T\leftrightarrow\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}\)
\(A:n\times n\) matrix, rank: \(r\), represents \(T:F^n\to F^n\) in standard basis
Important constraint: same basis for input and Output
Goal: Find invertible \(S\) s.t. \(S^{-1}AS\) is “simple”
Same vector space, same basis
Operators are typically used multiple times on a vector
\(Av\), \(A^2v\), \(A^3v\),\(\ldots\)
\(S^{-1}ASv\), \((S^{-1}AS)^2v=S^{-1}A^2Sv\),\(\ldots\)
Rest of course: study operators in above scenario