Symmetric matrices vector space property

Question

Let define $A=\{X\in\mathbb{C}^{3,3}:X=X^T\}$ where $U^T$ is a transpose of the matrix $U$. Then the $A$ is a vector space of symmetric matrices. Now how can we prove that $\{Y\in\mathbb{C}^{3,3}:\forall_{X\in A}\,YX\in A\}=\text{span}(I_3)$ where $I_{3}$ is the identity matrix.

My approach:

We can show that $\text{span}(I_3) \subseteq\{Y\in\mathbb{C}^{3,3}:\forall_{X\in A}\,YX\in A\}$ just by taking matrix from $\text{span}(I_3)$ for example $B=\begin{bmatrix} \alpha & 0 & 0 \\ 0 & \alpha & 0 \\ 0 & 0& \alpha \end{bmatrix}$ and then we see that $\forall_{X\in A}\,BX=\alpha I_{3}X=\alpha X$ so $BX\in A$ (because X was a symmetric matrix). So we now know that $B\in\{Y\in\mathbb{C}^{3,3}:\forall_{X\in A}\,YX\in A\}$ so $\text{span}(I_3) \subseteq\{Y\in\mathbb{C}^{3,3}:\forall_{X\in A}\,YX\in A\}$. But I have now idea how to continue this approach.

Answer 1

Let$$Y=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}.$$If $Y$ is such that $(\forall x\in A):YX\in A$, then in particular,$$Y.\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}\in A.$$But what this means is that\begin{bmatrix}a_{11}&0&0\\a_{21}&0&0\\a_{31}&0&0\end{bmatrix}is a symmetric matrix. In other words, $a_{21}=a_{31}=0$. Can you take it from here?

Answer 2

Assume that $Y$ is not in the span of $I_3$ e.g. $$Y=\begin{pmatrix}\alpha&0&0\\0&\beta&0\\0&0&\beta\end{pmatrix}$$ where $\alpha\neq\beta$ then $$Y\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}=\begin{pmatrix}0&\alpha&0\\\beta&0&0\\0&0&0\end{pmatrix}$$ is not symmetric and thus $Y\notin B$. You can proceed analogously in other cases.