<s>[INST] $\mathbf{A}^T = p(\mathbf{A})$, 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?
 [/INST] 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.
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