# Given a matrix $A$, is there a matrix $C$ with $AC = CA + A$?

Given this matrix A \begin{pmatrix}7+a&2&3&3+a\\2&7&7&11\\3&7&7&2\\3+a&11&2&11\end{pmatrix} where $a \in \mathbb{R}$

Is there a matrix $C \in \mathbb{R^{4x4}}$ with $AC = CA + A$ ?

Notes:

• $A$ is symmetric, and Hermitian
• I've thought of this $AC = CA + A \Rightarrow A = AC - CA$ (can we reach somewhere if we assume that $C = BAB^{-1}$ where $B$ is a regular matrix)
• $AC = CA + A \Rightarrow A= AC - CA$, if we assume that $C$ is the identity matrix then $CA = AC = I$, so $A = I - I \Rightarrow A = 0$, which is false, so there isn't a matrix $C$ (I am not sure)