Prove that matrix $A$ is invertible if $A$ is a polynomial with $A^{\top}$ as argument $A$ is a square matrix with elements in $\mathbb{R}$ such that $A = p(A^T) = p_n (A^T)^n + \ldots + p_1 A^T + p_0 \operatorname{Id}$, p is a polynomial, $p_i \in \mathbb{R}$, $p_0 \neq 0$. Prove that A is invertible. I am really stuck on that, tried to use the fact that $A = p(p(A))$ but had no success. 
 A: For the sake of contradiction assume that $A$ is not invertible. Then an $x\neq 0$ exists such that $Ax=0$. So,
$$0 = x^T A x = p_n x^T (A^T)^n x + \dots + p_0 x^T x = p_0 \lVert x \rVert$$
Since $p_0 \neq 0$ this implies $x=0$, which is a contradiction.
A: You might have $P(P(0)) = 0$, in this case your argument does not work.
Here, it is possible to say that, since $A$ and $A^T$ commutes : 
$$Ker(A) = Ker(A^T A) = Ker (A A^T) = Ker(A^T)$$
But it is easy to check that $Ker(A) \cap Ker(A^T) = \{0\}$. This concludes : you must have $Ker(A) = \{0\}$ and $A$ invertible.
EDIT : Here is more details.
$Ker(A) = Ker(A^T A)$ is always true. To show this : 


*

*$Ker(A) \subset Ker(A^T A)$ is clear.

*$Ker(A^T A) \subset Ker(A)$. Let $X \in Ker(A^T A)$. Then $X^T A^T A X = 0$. So $(AX)^T AX = 0$ and $AX = 0$. So $K \in Ker(A)$ (this trick might be clearer if you think with a scalar product).


This fact implies that, if $A$ is supposed normal, then $Ker(A) = Ker(A^T)$.
Here, $Ker(A) \cap Ker(A^T) = \{0\}$ comes from the polynomial relation between $A$ and $A^T$ (it is essential that $p_0 \neq 0$).
