Q:

If $\Lambda_A\in L(M_n),\;\Lambda_A(T)=AT$, show $\sigma\left(\Lambda_A\right)=\sigma (A)$

Let $A\in M_n$ and let $\Lambda_A\in L\left(M_n\right)$
$$\Lambda_A(T)=AT.$$
  Show $\sigma\left(\Lambda_A\right)=\sigma(A).$

Definition 1:
Let $V$ be a vector space over field $\mathbb F$ and $A\in L(V$). Scalar $\lambda_0\in\mathbb F$ is an eigenvalue of $A$ if $\exists x\in V,x\ne 0$ s.t. $Ax=\lambda_0 x$. Set $\sigma(A)$ of all eigenvalues of $A$ is called spectrum.

Definition 2:
Let $A\in L(V,W)$.
$$Ker A= A^{-1}\left(A\{0\}\right)=\{x\in V: Ax=0\}\leqslant V$$

Definition 3:
Let $E_A(\lambda_0)=\{x\in V:Ax=\lambda_0 x\}$. $E_A(\lambda_0)$ is the eigenspace of an eigenvalue $\lambda_0$. Note:
$$E_A(\lambda_0)=Ker(A-\lambda_0I).$$
$\lambda_0$ is an eigenvalue of $A$ $\iff$ $A$ is singular.

Definition 4:
Let $A\in M_n(\mathbb F)$. Polynomial $k_A(\lambda)\in\mathbb F[\lambda]$, $k_A=\det(A-\lambda_0I)$ is the characteristic polynomial of $A$.
Note: $\lambda_0$ is an eigenvalue of $A$ $\iff$ $k_A(\lambda_0)=0$,i.e., $\lambda_0$ is a root of the polynomial $k_A$.
$\deg(k_A)\leqslant n$

My attempt:
We have to prove: $$(\forall\lambda\in\mathbb F)\lambda\in\sigma\left(\Lambda_A\right)\implies\lambda\in\sigma(A)$$
$$\&$$
$$(\forall\lambda\in\mathbb F)\lambda\in\sigma(A)\implies\lambda\in\sigma\left(\Lambda_A\right)$$
i.e.
$$\sigma\left(\Lambda_A\right)\subseteq\sigma(A)\;\&\;\sigma(A)\subseteq\sigma\left(\Lambda_A\right)$$
For $\Lambda_A(T)$
$$AT=\lambda_0T\iff AT=\lambda_0 IT\iff(A-\lambda_0I)T=0$$
No matter how obvious this seems, I haven't come up to any justified step to prove both the matrix $A$ and the linear operator have to be singular. Everything seemed circular.
I know:
$$T\ne 0\nRightarrow \det(A-\lambda_0I)=0$$
because $T$ might be a zero-divisor and hence $\det T=0$.
Also, $\color{red}{\text{determinant isn't a linear form}}$, so I can't figure out how to end the proof.
What would be the next step? Thank you in advance!

A:

In general, given a matrix $B\in M_{n}$ whose columns are $(v_{1},...,v_{n})$, we have that the product $AB$ is the matrix whose columns are $(Av_{1},...,Av_{n})$. 
$\Rightarrow$ Assume $\lambda\in\sigma(A)$. Let $v$ be a non-zero eigenvector of $A$ corresponding to $\lambda$. Consider the matrix $B$ whose columns are $(v,\overset{\rightarrow}{0},\overset{\rightarrow}{0},...,\overset{\rightarrow}{0})$. Then, $\text{Λ}_{A}(B)=AB$ is the matrix whose columns are $(Av,\overset{\rightarrow}{0},\overset{\rightarrow}{0},...,\overset{\rightarrow}{0})$, i.e., it is $\lambda B$.
$\Leftarrow$ Assume $\lambda\in\sigma(\text{Λ}_{A})$. Let $B$ be a non-zero eigenvector of $\text{Λ}_{A}$ corresponding to $\lambda$. Let $v$ be a non-zero column of $B$. Then, $Av=\lambda v$, so that $\lambda\in\sigma(A)$. 

