Test diagonalizability of $T(A)=A^t$ over all $n\times n$ matrices

Question

Let $V$ be the vector space of all $n\times n$ matrices over $F$. Let $T$ be the linear operator on $V$ defined by $T(A)=A^t$. Test T for diagonalizability, and if $T$ is diagonalizable, find a basis for $V$ such that the matrix representation of T is diagonal.

My attempt: if c is the eigenvalue of $T$, $T(A)=cA$, $A^t=cA$ $\implies$ $(A^t)^t=(cA)^t$ $\implies$ $A=cA^t$ $\implies$ $A^t=cA=c(cA^t)=c^2A^t$ $\implies$ $A=c^2A$ $\implies$ $c^2=1$ $\implies$ $c=1,-1$.

So $A^t=A$ or $A^t=-A$, if $A^t=A$, then $A$ is a symmetric matrix, for example, a $3\times 3$ matrix:$\begin{bmatrix}0&1&0\\1&0&0\\0&0&0\end{bmatrix}$ satisfies $A^t=A$.

if $A^t=-A$, for example:$\begin{bmatrix}0&1&0\\-1&0&0\\0&0&0\end{bmatrix}$ satisfies $A^t=-A$.

But how do I find a basis? I guess that the answer is that it is diagonalizable.

Answer 1

Suppose $E_{ij}$ is an $n\times n$ matrix with its only non-zero entry at the intersection of row $i$ and column $j$ and this entry is $1$. $E_ {ij}$'s span $M_{n}(F)$. Now define $$ _sE_{ij}=(E_{ij}+E_{ij}^\top)/2,\quad_aE_{ij}=(E_{ij}-E_{ij}^\top)/2 $$ So $S=\{_sE_{ij},\;_aE_{ij}:i,j=1,2,\cdots, n\}$ is a spanning set of $M_n(F)$ with each of the matrix being an eigenvector of $T$. There are at most $2n^2$ matrices in $S$ but noting

• $_sE_{ij}={_s}E_{ji}$ lets us eliminate ${n^2\over2}-{n\over2}$ of these
• $_aE_{ij}=-{_a}E_{ji}$ lets us eliminate ${n^2\over2}-{n\over2}$ of these
• $_aE_{ii}=0I_n$ lets us eliminate $n$ of these

by virtue of linear dependence. So our basis is $$ \{_sE_{ij}:j\le i\le n, j=1,2,\cdots, n\}\cup\{_aE_{ij}:j< i\le n, j=1,2,\cdots, n\} $$ The first set has ${n^2\over2}+{n\over2}$ and the second set has ${n^2\over2}-{n\over2}$ elements.

Answer 2

Hint: Every matrix can be written as sum of a symmetrical matrix and an anti-simmetrical. Note the every simmetrical matrix is an eigenvector with eigenvalue $1$ and every anti-symmetrical matrix is an eigenvector with eigenvalue $-1$. Therefore $T$ is diagonalizable.

Answer 3

$\newcommand{\fvo}{1} \newcommand{\cdf}{\cdot & \cdot & \cdot & \cdot} \newcommand{\cdt}{\cdot & \cdot & \cdot} \newcommand{\cdo}{\cdot} \newcommand{\cdw}{\cdot & \cdot} $

$\newcommand{\vec}{\text{vec}}$ $\newcommand{\lb}{\left(}$ $\newcommand{\rb}{\right)}$ $\newcommand{\unity}{\bf\text{1}}$ Use the superoperator formalism to rewrite $A=A^T$ as

$$ \hat T \vec (A)=\vec(A)\\ $$ where $\hat T $ describes the Transposition Superoperator. For $4\times 4$ matrices this looks like

\begin{equation} \widehat T := \left(\, \begin{smallmatrix} \fvo & \cdt & \cdf & \cdf & \cdf \\ \cdf & \fvo & \cdt & \cdf & \cdf \\ \cdf & \cdf & \fvo & \cdt & \cdf \\ \cdf & \cdf & \cdf & \fvo & \cdt \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \cdo & \fvo & \cdw & \cdf & \cdf & \cdf \\ \cdf & \cdo & \fvo & \cdw & \cdf & \cdf \\ \cdf & \cdf & \cdo & \fvo & \cdw & \cdf \\ \cdf & \cdf & \cdf & \cdo & \fvo & \cdw \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \cdw & \fvo & \cdo & \cdf & \cdf & \cdf \\ \cdf & \cdw & \fvo & \cdo & \cdf & \cdf \\ \cdf & \cdf & \cdw & \fvo & \cdo & \cdf \\ \cdf & \cdf & \cdf & \cdw & \fvo & \cdo \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \cdt & \fvo & \cdf & \cdf & \cdf \\ \cdf & \cdt & \fvo & \cdf & \cdf \\ \cdf & \cdf & \cdt & \fvo & \cdf \\ \cdf & \cdf & \cdf & \cdt & \fvo \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\ \end{smallmatrix}\, \right) \end{equation}

This is a permutation matrix (so it is diagonalizable), with eigenvalues $c=\pm 1$...