Does $xA=x$ is true for any $x\ne 0\in\mathbb{F_q}^{1\times n}$ imply that $A=I$ where $A\in \mathbb{F_q}^{n\times n}$?

Question

Does $xA=x$ is true for any $x\ne 0\in\mathbb{F_q}^{1\times n}$ imply that $A=I$ where $A\in \mathbb{F_q}^{n\times n}$ and $\mathbb{F_q}$ is the finite field?

If it's true, how it can be shown formally?

Answer 1

Notice that taking $x=(0,\dots,0, 1, 0, \dots, 0)$ where the $1$ is in the $i$-th position, we have that $xA$ is equal to the $i$-th row of $A$. Therefore $xA=x$ implies that the $i$-th row of $A$ is $(0,\dots,0, 1, 0, \dots, 0)$. We can get this result for every row of $A$ and conclude that $A=I$.