# Eigenvalues of A are also eigenvalues of T

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?

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