Eigenvalues of a matrix $A$ and corresponding linear map (Linear algebra: Hoffman kunze 6.2.15) Let $V$ be the vector space of $n\times n$ matrices over the field $F$. 
Let $A$ be  a fixed $n\times n$ matrix over $F$.
let $T$ be a linear operator on $V$ defined as  $T(B) = AB$.
Question is to check whether :
$A$ and $T$ have same eigen values??? (Linear algebra: Hoffman kunze 6.2.15)
What i have done so far is :
let $\lambda \in F$ be an eigenvalue of $T$ i.e., $T(B)=\lambda B$ for some $B\neq 0$.
But, $T(B)=AB$ so, we have $AB=\lambda B$ i.e., $(A-\lambda I)B=0$
Now, $\det ((A-\lambda I).B)=0$ i.e., $\det (A-\lambda I).\det(B)=0$
Suppose $\det (A-\lambda I)\neq0$ this would imply $A-\lambda I$ is invertible and so $(A-\lambda I)B=0$ imply $(A-\lambda I)^{-1}(A-\lambda I)B=0$ i.e., $B=0$ (I really enjoyed this part... I am proud of myself here :P)
So, $\det(A-\lambda I)=0$ which imply that $\lambda$ is an eigenvalue of $A$ 
Conversely, Suppose $\lambda \in F$ is an eigenvalue of $A$.
I need to find some $B\neq 0$ such that $T(B)=\lambda B$
$T(B)=AB$ So, i have $AB=\lambda B$ i.e., $(A-\lambda I)B=0$
As $\lambda$ is an eigenvalue of $A$, $A-\lambda I$ is not invertible.
Now, Thanking an equivalent condition of if $AB\neq 0$ for any non zero matrix $B$ then $A$ is invertible , we see that there does exist $B\neq 0$ such that $(A-\lambda I)B=0$ and thus we are done.
Please suggest me if there are any gaps in the proof and if it can be simplified further.
Thank you
 A: I've had a cursory glance at your answer and it looks OK, but let me just mention one thing which may or may not make the situation clearer.
First note that if $A = \left[\begin{array}{c}a_1\\ \vdots\\ a_n\end{array}\right]$ where $a_i$ are the row vectors of $A$, and $B = [b_1\ \dots\ b_n]$ where $b_j$ are the columns vectors of $B$, then 
$$AB = \left[\begin{array}{c}a_1\\ \vdots\\ a_n\end{array}\right][b_1\ \dots\ b_n] = \left[\begin{array}{ccc}a_1b_1 & \dots & a_1b_n\\
\vdots & & \vdots\\ a_nb_1 & \dots & a_nb_n\end{array}\right] = [Ab_1\ \ \cdots\ \ Ab_n].$$
If $AB = \lambda B$, then $Ab_j = \lambda b_j$ for $1 \leq j \leq n$; that is, the columns of $B$ are eigenvectors of $A$ corresponding to the eigenvalue $\lambda$. Conversely, if $v$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$, then if we let $B= [v\ \ 0\ \ \cdots\ \ 0]$ we have 
$$AB = [Av\ \ A0\ \ \cdots\ \ A0] = [\lambda v\ \ 0\ \ \cdots\ \ 0] = \lambda[v\ \ 0\ \ \cdots\ \ 0] = \lambda B$$
so $B$ is an eigenvector of $T$ for the eigenvalue $\lambda$.
