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I need to find eigenvectors and eigenvalues of $T : M_{n\times n} \to M_{n\times n}$, $T(A) = A^T$ for $A\in M_{n\times n}$.

My intuition tells me that eigenvalues are $1$ and $-1$. $1$ for symmetric matrices and $-1$ for skew-symmetric matrices.

I can show that eigenvalue can only be $1$ if the entries on the diagonal are nonzero, from this I can say that if eigenvalue is $1$ then the matrix is symmetric.

For skew-symmetric matrices I can prove that eigenvalue is $-1$.

My problem for matrices which are not skew-symmetric and have only $0$ diagonal, how do I show that there is no eigenvalue ? In other words, how do I show that only eigenvalues possible is $-1$ and $1$ ?

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$\lambda$ is an Eigenvalue when $T(A)=\lambda A=A^T$.

Then $T(T(A))=(A^T)^T=A=\lambda^2A$.


The Eigenmatrices are such that either $A^T=A$ or $A^T=-A$, i.e. all symmetric or antisymmetric matrices.

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Hint: $(A^T)^T=A$ that is:

$$T^2=Id$$

Hence any eigenvalue must be $1$ or $-1$

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  • $\begingroup$ Are you saying that idempotent transformations can have only $-1$ and $1$ as eigenvalues ? $\endgroup$ – user8277998 Dec 12 '17 at 16:11
  • $\begingroup$ @123 no involution. I have corrected thanks $\endgroup$ – Guy Fsone Dec 12 '17 at 16:15

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