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I found out that the eigenvalues of this transformation is $\pm 1$. So for $\lambda =1$ the eigenvector is any symmetric matrix and for $\lambda = -1$ the eigenvector is any Skew-symmetric matrix. So what I don't know is how to formally define this eigenvectors can I say:

$v_1={A\in M_{n\times n}:A=A^t}$

$v_2={A\in M_{n\times n}:-A=A^t}$

Hamza
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Yudop
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    the dimension of the real vector space of square matrices is $n^2.$ What are the dimensions of your two subspaces, and what are bases for them? – Will Jagy Jul 10 '16 at 21:40

3 Answers3

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Let $A\in M_n(\mathbb{R})$ be a matrix so if we put $S=\frac{1}{2}(A+A^t)$ and $S'=\frac{1}{2}(A-A^t)$ then we have :

  1. $S^t=S$
  2. $S'^t=S'$
  3. $S+S'=A$

we can deduce easily that if we put $\mathcal{S}$ (resp $\mathcal{S}'$) the space of symmetric (resp Skew-symmetric) matrix : $$ M_n(\mathbb{R})=\mathcal{S}\oplus\mathcal{S}' $$

So let now the operator $$ \begin{array}{c} T: & M_n(\mathbb{R})& \to & M_n(\mathbb{R})\\ & A & \mapsto & A^t \end{array} $$ then $T$ is a linear operator and we need to see the eigenvalues of $T$ : $$ \lambda \in \sigma(T) \iff \exists A \in( M_n(\mathbb{R})-\{0\})\; \textrm { such that } T(A)=\lambda A $$ so it mean that : $$ A^t=\lambda A $$ two solution are obvious $\lambda=1,-1$, but using the fact that the sum of eigenspaces is a direct sum and the decomposition of our space we can see that $1$ and $-1$ are the only eigenvalues of $T$.

Community
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Hamza
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An eigenvector for the eigenvalue $\lambda$ is any nonzero vector $v$ satisfying $Tv = \lambda v$. Equivalently, it is a non-zero element of the eigenspace $\ker\{ T- \lambda I\} = \{ v \mid Tv = \lambda v \}$.

Adapted to your situation, this means that $A$ is an eigenvector for the eigenvalue $1$ iff it belongs to the set $\{A \in M_n \setminus \{ 0\}\mid A^t = A\}$ (and similarly for the other eigenvalue).

Alex Provost
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The subspaces (check they are subspaces) of $M_{n\times n}(\Bbb{R})$ defined as follows

$$\begin{align}S_n=&\{A\in M_{n\times n}(\Bbb{R}),\,A=A^T\}\\ A_n=&\{A\in M_{n\times n}(\Bbb{R}),\,A=-A^T\}\end{align}$$

have the zero matrix as intersection and have ${n(n+1)\over 2}$ and ${n(n-1)\over 2}$ respectively as dimensions and so they are supplementary. They correspond to the $\pm 1$ only eigenvalues.

Without using any dimension argument one can see that any matrix can be decomposed into a symmetric and an anti symmetric part

$$A={A+A^T\over 2}+{A-A^T\over 2}$$

marwalix
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