$\mathbf{A}^T = p(\mathbf{A})$, prove that $\mathbf{A}$ is invertible Let $\mathbf{A}$ be a square matrix defined over a field $\mathbb{R}$. It is known that $\mathbf{A}^\text{T} = p(\mathbf{A})$, where $p(\mathbf{A})$ is a polynomial with a constant coefficient $a_0 \neq 0$. 


*

*Prove that $\mathbf{A}$ is invertible. 

*Is it true that for every operator $\mathbb{\phi}: \mathbb{R}^n\rightarrow\mathbb{R}^n$ there exists some polynomial $p(x)$ and the basis for which the matrix $\phi$ satisfies the condition of $\mathbf{A}^\text{T} = p(\mathbf{A})$?


Solution
I.
By the definition of $\mathbf{A}$: $$\mathbf{A}^t = P(\mathbf{A}) = a_n \cdot \mathbf{A}^n + \cdots + a_1 \cdot \mathbf{A} + a_0 \cdot \mathbf{I}$$
Consider
$$\mathbf{A}^t \cdot \mathbf{A} = a_n \cdot \mathbf{A}^n \cdot \mathbf{A} + \cdots + a_1 \cdot \mathbf{A} \cdot \mathbf{A} + a_0 \cdot \mathbf{I} \cdot \mathbf{A} \\ = a_n \cdot \mathbf{A}^{n + 1} + \cdots + a_1 \cdot \mathbf{A}^2 + a_0 \cdot \mathbf{A}$$
$$\mathbf{A} \cdot \mathbf{A}^t = a_n \cdot \mathbf{A} \cdot \mathbf{A}^n + \cdots + a_1 \cdot \mathbf{A} \cdot \mathbf{A} + a_0 \cdot \mathbf{A} \cdot I \\ = a_n \cdot \mathbf{A}^{n + 1} + \cdots + a_1 \cdot \mathbf{A}^2 + a_0 \cdot \mathbf{A}$$
hence, $\mathbf{A}^t \cdot \mathbf{A} = \mathbf{A} \cdot \mathbf{A}^t$. Therefore, the matrix $\mathbf{A}$ is normal, which assumes $\text{ker}(\mathbf{A}) = \text{ker}(\mathbf{A^t})$. Assume that $\mathbf{A}$ is not invertible, then its kernel is non-trivial. Take non-zero $v$ from $\text{ker}(\mathbf{A})$:
$$\mathbf{A}^t \cdot v = a_n \cdot \mathbf{A}^n \cdot v + \cdots + a_1 \cdot \mathbf{A} \cdot v + a_0 \cdot \mathbf{I} \cdot v \\ \mathbf{A}^t \cdot v = 0 + \cdots + 0 + a_0 \cdot v = a_0 \cdot v$$
As per definition, $a_0 \neq 0$, so $a_0 \cdot v \neq 0$, and it follows that $\mathbf{A}^t \cdot v \neq 0$. However, the condition $\text{ker}(\mathbf{A})=\text{ker}(\mathbf{A^t})$ implies that $\mathbf{A}^t \cdot v = 0$. We have a contradiction, hence $\mathbf{A}$ is invertible.
II.
It was proven that $\mathbf{A}$ is normal if its transpose can be represented by a polynomial. But normality is independent of the basis chosen, so the second statement would mean that every operator is normal, which is not the case.
Any mistakes, improvements?

For the first question, suppose $Ax=0$. Then
$$
0=(Ax)^Tx=x^TA^Tx=x^T\left(p(A)x\right)=a_0x^Tx=a_0\|x\|_2^2.
$$
Therefore $x$ must be zero and $A$ is invertible.
