0

My question is about the answer to another question on this site -

Diagonalizable matrices that commute share eigenspace

I thought of leaving a comment but since it was a > 3 year old post, I thought I'd just ask a new question. In the answer provided by @user44197,

"Since the Eigen space of S is invariant under T and lets supposed $\lambda$ is an Eigen value of S with multiplicity m then,

$$M^{-1} T M = \left[ {\begin{array}{cc} T_{1,1} & 0 \\ 0 & T_{2,2} \\ \end{array} } \right] $$ "

Where $T_{1,1}$ is m x m.

I can't wrap my head around the reason for this. Why should the off-diagonal entries for $T$ become 0 when $M$ is a matrix that diagonalizes $S$.

Rohit Pandey
  • 6,803

1 Answers1

1

Your linked answer assumes that $M$ is chosen so that $D=M^{-1}SM$ is diagonal with grouped eigenvalues. That is, $$ D=\begin{bmatrix} \lambda_1 I_{m_1}&0&\cdots&0\\ 0&\lambda_2 I_{m_2}&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_k I_{m_k} \end{bmatrix} $$ where $\lambda_1,\ldots,\lambda_k$ are the distinct eigenvalues of $S$ with multiplicities $m_1,\ldots,m_k$. Let $T'=M^{-1}TM$, and write $T'$ in block form (with the same size blocks): $$ T'=\begin{bmatrix} T_{11}&T_{12}&\cdots&T_{1k}\\ T_{21}&T_{22}&\cdots&T_{2k}\\ \vdots&\vdots&\ddots&\vdots\\ T_{k1}&T_{k2}&\cdots&T_{kk} \end{bmatrix}. $$ Then $$ DT'=\begin{bmatrix} \lambda_1T_{11}&\lambda_1T_{12}&\cdots&\lambda_1T_{1k}\\ \lambda_2T_{21}&\lambda_2T_{22}&\cdots&\lambda_2T_{2k}\\ \vdots&\vdots&\ddots&\vdots\\ \lambda_kT_{k1}&\lambda_kT_{k2}&\cdots&\lambda_kT_{kk} \end{bmatrix}, $$ $$ T'D=\begin{bmatrix} \lambda_1T_{11}&\lambda_2T_{12}&\cdots&\lambda_kT_{1k}\\ \lambda_1T_{21}&\lambda_2T_{22}&\cdots&\lambda_kT_{2k}\\ \vdots&\vdots&\ddots&\vdots\\ \lambda_1T_{k1}&\lambda_2T_{k2}&\cdots&\lambda_kT_{kk} \end{bmatrix}. $$ But $$ DT'=M^{-1}STM=M^{-1}TSM=T'D. $$ Since $\lambda_i\neq\lambda_j$ for $i\neq j$, we have $T_{ij}=0$ for $i\neq j$.

stewbasic
  • 6,131
  • Its clear now - one point of confusion I still have - if all the $m_i$ are 1. This seems to imply that $T'$ will be diagonal as well? That can't be right. – Rohit Pandey May 24 '17 at 01:16
  • That is right. If $S$ and $T$ commute then $T$ must preserve the eigenspaces of $S$. If all eigenvalues of $S$ are distinct then those eigenspaces are 1-dimensional, so $T$ must act as a (possibly different) scalar on each eigenspace. – stewbasic May 24 '17 at 01:26