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Let $A$ be real square matrix, satisfying $A^T=p(A)$ for some polynomial $$p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$ such that $a_0\neq 0$.

I have to prove that $A$ is invertible and I have no idea how to do it.

Adam Chalumeau
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Hrackadont
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2 Answers2

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Hint. Suppose $Ax=0$. Now consider the quantity $x^TA^Tx$.

user1551
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Since $A^T=p(A)$ commutes with $A$, we can simultaneously unitarily upper-triangularise $A,A^T$, hence diagonalise $A,A^T$ (using $A^T=p(A)$). So the eigenvalues of $A$ satisfies $\bar\lambda=p(\lambda)$, hence $\lambda\neq 0$ (since $a_0\neq 0$).

user10354138
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    You don't necessarily have $A = A^T$. For example, if $A = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$ then $A^T = A^3$ but $A$ is not symmetric. – Daniel Schepler Jun 25 '19 at 20:56
  • That two matrices commute doesn't necessarily mean they can be (simultaneously or not( diagonalizable. There are commuting non-diagonalizable matrices... – DonAntonio Jun 25 '19 at 21:00
  • The same example I gave before shows $A$ doesn't necessarily have to be diagonalizable as a real matrix: so this would need to go into complex territory. Then, you might need to be careful about how $A^T$ translates. My thought is that if $\lambda$ is a complex eigenvalue of $A$ then $\bar \lambda = p(\lambda)$ - using the representation theorem that a normal real matrix is orthogonally similar to a block-diagonal matrix with diagonal blocks of the form $\begin{bmatrix} \lambda \end{bmatrix}$ and $\begin{bmatrix} a&-b\b&a \end{bmatrix}$. – Daniel Schepler Jun 25 '19 at 21:08
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    Yes I made a typo. Meant $\bar\lambda=p(\lambda)$ instead of $\lambda=p(\lambda)$. It is immediate from $A^\dagger=p(A)$ without needing representation theorem of normal matrices (we are given that for the original $A$). – user10354138 Jun 25 '19 at 21:12