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

1. ## 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!

2. Accidental post

3. 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.

4. 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.

5. 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.

6. 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!

7. 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 $\lambda^2=1$.

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