2
$\begingroup$

Good morning can someone help me in this exercise.

Let $A\in M(n,\mathbb{R})$ for wich exists an integer $k\ge 1$ such that $A^t=A^k$ where $A^t$ is the trasposed matrix, let also $J(A)$ the jordan canonical form of $A$. Prove that $J(A)^k=J(A)^t$ and determinate all possible jordan canonical form beetwen the matrix such that $A^2=A^t$.

For the first question i managed to prove that $J(A)^k$ and $J(A)^t$ are similar infact $A$ is similar to $J(A)$ so exists an invertible matrix $M$ such that $A=M^{-1}J(A)M$ so $A^k=M^{-1}J(A)^kM$ and $A^t=M^tJ(A)^t(M^{-1})^t$ but $A^t=J(A)$ so $M^{-1}J(A)^kM=M^tJ(A)^t(M^{-1})^t$ that prove that $J(A)^k$ is similat to $J(A)^t$ but i don't know how to conclude that they are equal.

$\endgroup$
  • $\begingroup$ I may lack intuition but I would say that $J(A)^T$ is not a Jordan form unless $J(A)$ is actually diagonal, because the ones are on the wrong side of the diagonal then. Speculation : The eigenvalues of $A^T$ may be the same as for $A$ in your case which means we can express $J(A^T)=J(A)$. $\endgroup$ – P. Quinton Sep 3 '18 at 18:08
  • 1
    $\begingroup$ @P.Quinton The eigenvalues of $A$ and $A^T$ are always the same; even better $A$ and $A^T$ have the same characteristic polynomial, minimal polynomial etc. and so they have the same Jordan normal form (up to the limits of uniqueness of Jordan normal forms). $\endgroup$ – Kusma Sep 3 '18 at 19:59
1
$\begingroup$

From $A^k=A^t$ we see that $A^tA=A^{k+1}=AA^t$. Hence $A$ is normal, and diagonalizable over $\mathbb{C}$. Hence $J(A)^t=J(A)=J(A^t)=J(A^k)=J(A^k)^t$. $J(A^k)$ and $(J(A))^k$ are only equal if you decide to order the eigenvalues in the corresponding way, and I am not sure there is a canonical way to do that.

In any case, your $A^2=A^t$ question is reduced to the same question for diagonal matrices.

$\endgroup$
  • $\begingroup$ Thank you very much you have been really clear. $\endgroup$ – Tyrion Sep 4 '18 at 14:43
  • $\begingroup$ @Tyrion: Note that you can accept an answer if it helped, by clicking on the check mark. See math.stackexchange.com/help/someone-answers for more information. $\endgroup$ – Martin R May 13 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.