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?
Best Answer
Your proof for part 1 is perfect.
For part 2, I assume that the question should be as follows:
First, to correct the answer you gave, it is not true that normality is independent of the basis chosen: If you apply a non-unitary change of basis to a normal matrix, then the resulting matrix might not be normal. For instance, note that if we take $$ A = \pmatrix{0&2\\2&0}, \quad S = \pmatrix{2&0\\0&1}, $$ then $A$ is normal but $SAS^{-1}$ is not.
One way to answer the question is to give an example of an operator for which the condition cannot hold. Consider in particular the operator $\phi(x) = Mx$ with $$ M = \pmatrix{0&1\\0&0}. $$ $\phi$ fails to be diagonalizable and therefore cannot be normal relative to any choice of basis. Thus, it cannot be the case that the matrix $A$ of $\phi$ satisfies $A^\top = p(A)$.