The eigenvectors of the transpose operator

Question

Define $T : M_{n×n}(\mathbb{R}) → M_{n×n}(\mathbb{R})$ by $T(A) := A^t$.

I know that the corresponding eigenvalues are $+1$ and $-1$, but I'm not sure how to find the eigenvectors of this transformation, in the case of a $2\times 2$ matrix it's simple, but not in the general case.

Answer 1

$$A\text{ symmetric }\Rightarrow T(A)=A^T=A=1A\Rightarrow A \text{ is eigen value associated to }\lambda_1=1,$$ $$A\text{ skew-symmetric }\Rightarrow T(A)=A^T=-A=(-1)A\Rightarrow A \text{ is eigen value associated to }\lambda_2=-1.$$

On the other hand, $M_{n×n}(\mathbb{R})$ is direct sum of the subspaces of symmetric and skew-symmetric matrices, so there are no more eigenvalues.

Answer 2

HINT: If the eigenvalue is $1$ means that $A=A^t$, so how are called this type of matrices? And if the eigenvalue is $-1$ means that $A=-A^t$, so again: who are these matrices?

Answer 3

The vector space $\frak{M}$ of $n \times n$ matrices is the direct sum of the (sub)vector spaces of symmetric and skew-symmetric (anti-symmetric) matrices:

$$\frak{M}=\frak{S} \oplus \frak{A}$$

with dim$(\frak{S})$=$\dfrac{n(n+1)}{2}$ and dim$(\frak{A})$=$\dfrac{n(n-1)}{2}.$

(see for example (Direct summand of skew-symmetric and symmetric matrices))

In order to be more specific, let us take the case $n=3$ (the general case can be easily understood through this example). In this case dim$(\frak{S})$=$6$ and dim$(\frak{A})$=$3.$

A simple basis of eigenvectors of the transpose operator $T:A\mapsto A^T$ associated with

• eigenvalue 1 is (symmetric matrices)

$$\pmatrix{1&0&0\\0&0&0\\0&0&0}, \pmatrix{0&0&0\\0&1&0\\0&0&0},\pmatrix{0&0&0\\0&0&0\\0&0&1}$$

$$\pmatrix{0&1&0\\1&0&0\\0&0&0}, \pmatrix{0&0&1\\0&0&0\\1&0&0},\pmatrix{0&0&0\\0&0&1\\0&1&0}$$

• eigenvalue -1 is (skew-symmetric matrices)

$$\pmatrix{0&1&0\\-1&0&0\\0&0&0}, \pmatrix{0&0&1\\0&0&0\\-1&0&0},\pmatrix{0&0&0\\0&0&1\\0&-1&0}$$