$Z_A := \{B \in \mathbb{K}^{n,n}:AB = BA\}$ A,B matrices. Proof $Z_A$ is a $\mathbb{K}$-subalgebra of $\mathbb{K^{n,n}}$.

Question

Let $Z_A := \{B \in \mathbb{K}^{n,n}:AB = BA\}, n \in \mathbb{N^*}, \mathbb{K}$ a field and A, B be the set of muliplicative interchangable matrices. Proof, that $Z_A$ is a $\mathbb{K}$-subalgebra of $\mathbb{K^{n,n}}$.

The only property Im failing to prove is, that if you multiply A and B that the solution is in $Z_A$. In other words that if you multiply two multiplicative interchangable matrices you get another one.

I found out that for commutative matrices there exist common Eigenvalues/Eigenvectors, that might be of help.

Answer 1

You are given a fixed $A \in \mathbb{K}^{n \times n}$, then you want to show that the set \begin{align} Z_A:= \{ B \in \mathbb{K}^{n \times n} \mid AB =BA \} \subset \mathbb{K}^{n \times n} \end{align} is a subalgebra of $\mathbb{K}^{n \times n}$.
Since $\mathbb{K}^{n \times n}$ is an algebra, you only have to show that $Z_A$ is closed under addition, multiplication and scalar multiplication.
Thus if you take $B,C \in Z_A$ and $\lambda \in \mathbb{K}$, you must show that $B+C,BC,\lambda B \in Z_A$.
Thus you must show that $a(B+C)=(B+C)A$, $a(BC)=(BC)A$ and $A(\lambda B)=(\lambda B)A$.

We now show that $BC \in Z_A$ by using the facts that $AB=BA$, $AC = CA$ and that the multiplication is associative, thus it does not matter if we multiply $A$ with $BC$ to obtain $A(BC)$ or $AB$ with $C$ to obtain $(AB)C$, as they are the same. Then we get \begin{align} A(BC)=(AB)C=(BA)C=B(AC)=B(CA)=(BC)A. \end{align} So we see that $A(BC) =(BC)A$, and thus that $BC \in Z_A$.

Answer 2

You do not need eigenvectors for this, just computation. Let $B, C \in Z_A$, we will show $(BC)A = A(BC)$, that is $BC \in Z_A$: $$ (BC)A = B(CA) = B(AC)= (BA)C= (AB)C = A(BC). $$