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

• May 13th 2010, 01:34 AM
kaelbu
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!
• May 13th 2010, 06:12 AM
HallsofIvy
Accidental post
• May 13th 2010, 06:15 AM
HallsofIvy
T is a linear operation on matrices that maps a matrix into its transpose? I started to say "Think about the fact that $A$ and $A^T$ have the same main diagonal", but then it occured to me that A might not be a square matrix. Certainly, you can look at $a_{11}$ and $a_22$ which must be the same in both matrices.
• May 13th 2010, 07:34 AM
Opalg
Quote:

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.

Hint: $T^2$ is the identity.
• May 13th 2010, 08:21 AM
kaelbu
Quote:

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

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 $M_(nxn)(F)$).
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.
• May 13th 2010, 08:36 AM
kaelbu
Oh!
If T^2 is the identity, T must also be the identity, I think. Oh no, each diagonal entry must be the $\sqrt(1) = + or - 1$.
Thanks! You're the greatest!
• May 13th 2010, 11:39 AM
Opalg
Quote:

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 $\sqrt(1) = + or - 1$.
Thanks! You're the greatest!

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

T is defined by $T(A) = A^{\textsc{t}}$. So $T^2(A)= (A^{\textsc{t}})^{\textsc{t}} = A$. And $\lambda$ is an eigenvalue of T if there is a nonzero matrix A such that $T(A) = \lambda A$. Then $A = T^2(A) = T(T(A)) = T(\lambda A) = \lambda T(A) = \lambda^2A$. So $(\lambda^2-1)A = 0$ and hence $\lambda^2=1$.
• May 13th 2010, 04:43 PM
kaelbu
That's more or less what I meant, though I did it slightly differently.
Thanks again for the help.