Subspace of Set of n by n Matrices Proof

Question

I am wondering whether this proof of a set being a vector space is valid. I am allowed to assume that $M_{n,n}$ is a vector space over the field $\mathbb{R}$, and basic properties of matrices.

let A be a fixed matrix $\in M_{n,n}$. Show that $S = \{X\in M_{n,n}: AXA = X\}$ is a vector space.

$\underline{\text{Proof}}$

We assume that $M_{n,n}$ is a vector space over the field $\mathbb{R}$, and the basic properties of matrices.

We know that $S \subset M_{n,n}$ by definition of $S$, and so it suffices to show that $S$ is a subspace of $M_{n,n}$.

Now, consider the equation \begin{align} A\mathbf{0}A &= (A\mathbf{0})A \\ &= \mathbf{0}A \\ &= \mathbf{0} \end{align} by associativity of matrix multiplication and properties of the n by n zero matrix.

Hence, $\mathbf{0} \in S$. Therefore $S$ is non-empty.

Now, let $X,Y \in S$. This implies that \begin{align} AXA &= X \\ AYA &= Y \end{align}

Hence, $$X + Y = AXA + AYA = A(XA + YA) = A(X + Y)A$$ This implies that $X + Y \in S$. Therefore $S$ is closed under matrix addition.

Now, let $\alpha \in \mathbb{R}$ and $X \in S$. This implies that \begin{align} AXA &= X \\ \Rightarrow \alpha(AXA) &= \alpha X \\ \Leftrightarrow A(\alpha X)A &= \alpha X \end{align}

Hence, $\alpha X \in S$. Therefore $S$ is closed under scalar multiplication.

Hence, by the Subspace Theorem, S is a subspace of $M_{n,n}$. Therefore $S$ is a vector space. This completes the proof.

I am not so sure about this step: $\alpha(AXA) = \alpha X \\ \Leftrightarrow A(\alpha X)A = \alpha X$

Would this be considered a basic property of matrices that can be assumed?

Edit: My question has been resolved.

Answer 1

Yes$$\alpha(AXA) = A(\alpha X)A$$

Therefore $$\alpha(AXA) = \alpha X \\ \Leftrightarrow A(\alpha X)A = \alpha X$$ Is correct.

Answer 2

Yes it is. It is just an element-wise multiplication of a matrix with a scalar, or if you are still confused you can argue about it as $\alpha A X A = \alpha A \left( \frac{1}{\alpha} \cdot \alpha\right) X A = \alpha \cdot \frac{1}{\alpha} \cdot A \cdot \alpha X A $ since right and left multiplcation by scalar is equivalent.