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?

up vote 6 down vote accepted

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 $\lambda$.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.