Andrew Thangaraj
Aug-Nov 2020
\(S:V\to W\), \(T:W\to V\)
\(ST:W\to W\) and \(TS: V\to V\) are operators
If \(\lambda\ne0\) is an eigenvalue of \(ST\), then it is an eigenvalue of \(TS\) as well.
Proof
\(v\): \(STv=\lambda v\), \(\lambda\ne0\)
Then, \(Tv\ne0\),
and \(\lambda Tv=T(\lambda v)=T(STv)=(TS)Tv\)
So, \(Tv\): eigenvector of \(TS\) with eigenvalue \(\lambda\)
\(A\): \(m\times n\) and \(B\): \(n\times m\)
\(A\), \(B\): represent suitable linear maps
\(AB\): \(m\times m\), \(BA\): \(n\times n\)
\(t^n\text{det}(tI_m-AB)=t^m\text{det}(tI_n-BA)\)
Proof
\(C=\begin{bmatrix} tI_m&A\\ B&I_n \end{bmatrix}\), \(D=\begin{bmatrix} I_m&0\\ -B&tI_n \end{bmatrix}\)
det\((CD)=\) det\((DC)\)
det\(\begin{bmatrix} tI_m-AB&tA\\ 0&tI_n \end{bmatrix}=\) det\(\begin{bmatrix} tI_m&A\\ 0&tI_n-BA \end{bmatrix}\)
\(n>m\)
\(\{\)Eigenvalues of \(BA\}=\{\)Eigenvalues of \(AB\}\cup(n-m)\) zeros
\(V\): finite-dimensional inner product space
\(T:V\to V\), an operator
\(T^*:V\to V\), adjoint of \(T\)
Properties
\(\langle Tu,v\rangle=\langle u,T^*v\rangle\)
null \(T^*=\) \((\)range \(T)^{\perp}\)
dim range \(T=\) dim range \(T^*\)
dim null \(T=\) dim null \(T^*\)
\(T:V\to V\), an operator
if \(\lambda\) is an eigenvalue of \(T\), \(\bar{\lambda}\) is an eigenvalue of \(T^*\)
Proof
dim range \((T-\lambda I)<\) dim \(V\)
\((T-\lambda I)^* = T^*-\bar{\lambda}I\)
Since dim range \((T-\lambda I)=\) dim range \((T^*-\bar{\lambda}I)\),
dim range \((T^*-\bar{\lambda}I)<\) dim \(V\)
\(T:V\to W\), \(T^*:W\to V\)
\(TT^*: W\to W\), \(T^*T: V\to V\)
Nonzero eigenvalues of \(TT^*\) and \(T^*T\) are the same
Differences only in additional zero eigenvalues for one of the two
Let dim \(V\ge\) dim \(W\)
Singular values of a linear map \(T\): eigenvalues of \(TT^*\)
\(T:V\to W\), linear map and \(w\in W\)
Least squares: Solve \(\min\limits_v\,\lVert Tv-w\rVert\)
Solution: \(Tv-w\) is orthogonal to range \(T\), or
\(Tv-w\in\) \((\)range \(T)^{\perp}\)
Since null \(T^*=\) \((\)range \(T)^{\perp}\), we require
\(T^*(Tv-w)=0\)
Normal equation: \(T^*Tv=T^*w\)
Suppose \(T^*T\) is invertible
Pseudo-inverse of \(T\): \((T^* T)^{-1}T^*\)
Why? \(((T^* T)^{-1}T^*)T=I\)