Prove that matrix $A\in \mathbb{R}^{n \times n}$ is invertible if $A^T = p(A)$

Question

I have to prove that

a matrix $A\in \mathbb{R}^{n \times n}$ is invertible if $A^T = p(A)$ where $p(A)$ is a polynomial with non-zero last coefficient.

I've tried to use that if $A^T=p(A)$ then $A=p(A^T)$ and to look at $$AA^T = (a_0E+a_1A^T+...+a_m(A^T)^m)(a_0E+a_1A+...+a_mA^m)$$ and somehow to get the expression like $cE=A(...)$ or $cE=A^T(...)$ which will prove what I need but failed.

Answer 1

If $A^T=p(A)$, then $A=p(A^T)=p(p(A))$. Since $p$ and therefore $p\circ p$ has a non-zero constant term, we can rearrange $A=p(p(A))$ to have the constant term (times the identity matrix) alone on one side, and a multiple of $A$ on the other side. Divide by the constant term, and you have something of the form $I=A\cdot q(A)$.

Alternately, $p(p(A))-A=0$ implies that $p(p(x))-x$ must be a multiple of the minimal polynomial of $A$. Since the minimal polynomial thus has a non-zero constant term, $A$ can't have $0$ as an eigenvalue.

Answer 2

If $p$ has a non zero last coefficient, then we can write $p(x) = x q(x) + k$.

Then $A^T = A q(A) + kI$ and if $Av = \lambda v$ for some unit vector $v$, we have $v^T A^T v = \lambda = v^T ( A q(A) + kI) v = \lambda q(\lambda) + k$ from which we see that $\lambda \neq 0$.