Eigenvectors and Values for T(A)=A^t

I'm having some trouble getting started with this, if someone could point me in the right direction that would awesome.
If T(A) = A^t (A-transpose) show that +1 and - 1 are the only eigenvalues of T.
I think that my problem is that I don't know how to symbolize the matrix of linear transformation for T, but perhaps there is another way to go about doing this problem.
Thanks!

Accidental post

T is a linear operation on matrices that maps a matrix into its transpose? I started to say "Think about the fact that $\displaystyle A$ and $\displaystyle A^T$ have the same main diagonal", but then it occured to me that A might not be a square matrix. Certainly, you can look at $\displaystyle a_{11}$ and $\displaystyle a_22$ which must be the same in both matrices.

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Originally Posted by kaelbu

I'm having some trouble getting started with this, if someone could point me in the right direction that would awesome.
If T(A) = A^t (A-transpose) show that +1 and - 1 are the only eigenvalues of T.
I think that my problem is that I don't know how to symbolize the matrix of linear transformation for T, but perhaps there is another way to go about doing this problem.

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Hint: $\displaystyle T^2$ is the identity.

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Originally Posted by HallsofIvy

I started to say "Think about the fact that $\displaystyle A$ and $\displaystyle A^T$ have the same main diagonal", but then it occured to me that A might not be a square matrix.

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Isn't it true that for a matrix to have transpose it must be square?
Also, I the fact that A is a square matrix was actually part of the problem (T is a linear operator on $\displaystyle M_(nxn)(F)$).
Sorry I did not include that in the original problem, I guess I have a bad habit of over looking that part of the problem because I don't really have any idea what's going on, therefore it doesn't really make any difference to me what vector space the transformations in.

Oh!
If T^2 is the identity, T must also be the identity, I think. Oh no, each diagonal entry must be the $\displaystyle \sqrt(1) = + or - 1 $.
Thanks! You're the greatest!

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Originally Posted by kaelbu

Oh!
If T^2 is the identity, T must also be the identity, I think. Oh no, each diagonal entry must be the $\displaystyle \sqrt(1) = + or - 1 $.
Thanks! You're the greatest!

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Thanks for the compliment, but I'm not sure what you mean by the diagonal entries of T?

T is defined by $\displaystyle T(A) = A^{\textsc{t}}$. So $\displaystyle T^2(A)= (A^{\textsc{t}})^{\textsc{t}} = A$. And $\displaystyle \lambda$ is an eigenvalue of T if there is a nonzero matrix A such that $\displaystyle T(A) = \lambda A$. Then $\displaystyle A = T^2(A) = T(T(A)) = T(\lambda A) = \lambda T(A) = \lambda^2A$. So $\displaystyle (\lambda^2-1)A = 0$ and hence $\displaystyle \lambda^2=1$.

That's more or less what I meant, though I did it slightly differently.
Thanks again for the help.