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

May 2010
13
3
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!
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Accidental post
 
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HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
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: \(\displaystyle T^2\) is the identity.
 
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May 2010
13
3
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.
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.
 
May 2010
13
3
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!
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
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!
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 \(\displaystyle \lambda^2=1\).
 
May 2010
13
3
That's more or less what I meant, though I did it slightly differently.
Thanks again for the help.