Andrew Thangaraj
Aug-Nov 2020
\(T:V\to V\), diagonalizable
Eigenvector basis of \(V\): \(\{v_1,\ldots,v_n\}\)
\(Tv_i=\lambda_iv_i\)
Matrix of \(T\) w.r.t. eigenvector basis
\(D=\begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}\) (diagonal)
\(A\): \(n\times n\) matrix, represents \(T\) in standard basis
Matrix with \(v_i\) (coordinates in standard basis) as \(i\)-th column
\(S=\begin{bmatrix} \vdots&\cdots&\vdots&\cdots&\vdots\\ v_1&\cdots&v_i&\cdots&v_n\\ \vdots&\cdots&\vdots&\cdots&\vdots \end{bmatrix}\)
Let \(S^{-1}=\begin{bmatrix} \vdots&\vdots&\vdots\\ \cdots&u^T_i&\cdots\\ \vdots&\vdots&\vdots \end{bmatrix}\), matrix with \(u^T_i\) as \(i\)-th row
\(A=SDS^{-1}=\begin{bmatrix} \vdots&\cdots&\vdots&\cdots&\vdots\\ v_1&\cdots&v_i&\cdots&v_n\\ \vdots&\cdots&\vdots&\cdots&\vdots \end{bmatrix}\begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}\begin{bmatrix} \vdots&\vdots&\vdots\\ \cdots&u^T_i&\cdots\\ \vdots&\vdots&\vdots \end{bmatrix}\)
\(A=\lambda_1v_1u^T_1+\cdots+\lambda_nv_nu^T_n\)
\(T:V\to V\): diagonalizable
There exists a basis s.t. matrix of \(T\) has the following form:
\(A=\lambda_1v_1u^T_1+\cdots+\lambda_nv_nu^T_n\)
\(v_i\): eigenvectors in the basis, \(u_i\): rows of inverse matrix
Suppose rank \(T=r\).
There exists a basis s.t. matrix of \(T\) has the following form:
\(A=\lambda_1v_1u^T_1+\cdots+\lambda_rv_ru^T_r\)
\(\lambda_i\): nonzero eigenvalues with corresponding eigenvectors \(v_i\), \(1\le i\le r\)
\(A=\begin{bmatrix} 1&0\\ 0&2 \end{bmatrix}\) (self-adjoint, diagonal)
\(A=\begin{bmatrix} 1\\ 0 \end{bmatrix}\begin{bmatrix} 1&0 \end{bmatrix}+2\begin{bmatrix} 0\\ 1 \end{bmatrix}\begin{bmatrix} 0&1 \end{bmatrix}\)
\(A=\begin{bmatrix} 1&2\\ 3&2 \end{bmatrix}\) (not self-adjoint)
\(\lambda=-1,4\)
\(S=\begin{bmatrix} 1&2\\ -1&3 \end{bmatrix}\), \(S^{-1}=\begin{bmatrix} 3/5&-2/5\\ 1/5&1/5 \end{bmatrix}\)
\(A=-\begin{bmatrix} 1\\ -1 \end{bmatrix}\begin{bmatrix} 3/5&-2/5 \end{bmatrix}+4\begin{bmatrix} 2\\ 3 \end{bmatrix}\begin{bmatrix} 1/5&1/5 \end{bmatrix}\)
\(A=\begin{bmatrix} 2&-3\\ 3&2 \end{bmatrix}\) (normal)
\(\lambda=2+3i,2-3i\)
\(S=\frac{1}{\sqrt{2}}\begin{bmatrix} i&-i\\ 1&1 \end{bmatrix}\), \(S^{-1}=\frac{1}{\sqrt{2}}\begin{bmatrix} -i&1\\ i&1 \end{bmatrix}\)
\(A=\frac{2+3i}{2}\begin{bmatrix} i\\ 1 \end{bmatrix}\begin{bmatrix} -i&1 \end{bmatrix}+\frac{2-3i}{2}\begin{bmatrix} -i\\ 1 \end{bmatrix}\begin{bmatrix} i&1 \end{bmatrix}\)
Observations
Eigenvector basis: Orthonormal
\(V\): inner product space over \(C\), \(T:V\to V\)
The following are equivalent.
- \(T\) is normal.
- \(V\) has an orthonormal basis of eigenvectors of \(T\).
- \(T\) is diagonal w.r.t. an orthonormal basis.
Proof
Clearly, (2) implies (3), and (3) implies (2)
Proof of (3) implies (1)
\(B\): orthonormal basis s.t. \(M(T,B)\) is diagonal
Then, \(M(T^*,B)\): conjugate-transpose of \(M(T,B)\), which is diagonal
So, \(M(T,B)M(T^*,B)=M(T^*,B)M(T,B)\), or \(T\) commutes with \(T^*\)
Proof of (1) implies (3)
Schur’s theorem: \(T\) is upper-triangular w.r.t. an orthonormal basis
Orthonormal basis: \(\{e_1,\cdots,e_n\}\)
\(M(T,B)=\begin{bmatrix} a_{11}&\cdots&a_{1n}\\ &\ddots&\vdots\\ 0 & &a_{nn} \end{bmatrix}\), \(M(T^*,B)=\begin{bmatrix} \overline{a_{11}}& &0\\ \vdots &\ddots& \\ \overline{a_{1n}}&\cdots&\overline{a_{nn}} \end{bmatrix}\)
We will show that \(M(T,B)\) is, in fact, diagonal
\(\lVert Te_1\rVert^2=\lvert a_{11}\rvert^2=\lVert T^*e_1\rVert^2=\lvert a_{11}\rvert^2+\cdots+\lvert a_{1n}\rvert^2\)
So, \(a_{12}=\cdots=a_{1n}=0\)
\(\lVert Te_2\rVert^2=\lvert a_{22}\rvert^2=\lVert T^*e_2\rVert^2=\lvert a_{22}\rvert^2+\cdots+\lvert a_{2n}\rvert^2\)
So, \(a_{23}=\cdots=a_{2n}=0\)
and so on…
\(A\): \(n\times n\) matrix, represents a normal operator \(T\) in standard basis
\(T\) is diagonal w.r.t. orthonormal eigenvector basis \(\{e_1,\cdots,e_n\}\)
Corresponding eigenvalues: \(\{\lambda_1,\ldots,\lambda_n\}\)
Matrix with \(e_i\) (coordinates in standard basis) as \(i\)-th column
\(S=\begin{bmatrix} \vdots&\cdots&\vdots&\cdots&\vdots\\ e_1&\cdots&e_i&\cdots&e_n\\ \vdots&\cdots&\vdots&\cdots&\vdots \end{bmatrix}\)
\(S^{-1}=\begin{bmatrix} \vdots&\vdots&\vdots\\ \cdots&\overline{e^T_i}&\cdots\\ \vdots&\vdots&\vdots \end{bmatrix}\), matrix with \(\overline{e^T_i}\) as \(i\)-th row
\(A=SDS^{-1}=\lambda_1e_1\overline{e^T_1}+\cdots+\lambda_n e_n \overline{e^T_n}\)