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.

  • $\begingroup$ 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
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    $\begingroup$ @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

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.

  • $\begingroup$ 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

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