Eigenvalues of A are also eigenvalues of T

Question

Let $V$ be the set of all $n\times n$ matrices over a field $F$. Let $A$ be a fixed element of $V$. Define a linear operator $T$ on $V$ by $T(B)=AB$. I am trying to show that if $\lambda$ is an eigenvalue of $A$, then $\lambda$ is also an eigenvalue of $T$.

So suppose $Av=\lambda v$ for some $v\neq 0$ in $V$ and $\lambda\in F$. So I'd like to prove the existence of a matrix $B$ such that $T(B)=AB=\lambda A$, or equivalently, show that $T-\lambda I_V$ is not invertible (or injective or surjective). But I am not sure how to proceed from here. What can I do?

Answer 1

Take the matrix such that all of its columns are equal to $v$.

Answer 2

An eigenvalue of $T$ is some number $\lambda$ such that $T(B)=\lambda B$. Thus, we need to find a matrix $B$ such that $AB=\lambda B$, not $AB=\lambda A$. We can write a matrix as a row vector of column vectors, i.e. $B=[b_1,b_2,b_3...]$. Matrix multiplication acts on those columns independently: $AB=[Ab_1,Ab_,Ab_3...]$. If $b_i=v$ for all $i$, then $AB=[\lambda v, \lambda v, \lambda v, ...]=\lambda [v,v,v,...]=\lambda B$. So this shows that $[v,v,v,...]$ is an eigenvector with eigenvalue $\lambda$.