4 x 4 Characteristic Values For $V = \mathbb{R}^{4 \times 4}$ the vector space of $4 \times 4$ matrices over $\mathbb{R}$, we're given that the matrix $$B = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 3 & 0 \\
0 & 0 & 0 & 4 \\
\end{bmatrix}$$ and we define $T_B: V \to V$ by $T_B(A) = BA - AB$ for each $A \in V$. We want to find all eigenvalues of $T_B$ and the dimension of the associated eigenspace to each eigenvalue. I've found the matrix of $xI - (BA - AB)$ to be $$\begin{bmatrix}
x & A_{12} & 2A_{13} & 3A_{14} \\
-A_{21} & x & A_{23} & 2A_{24} \\
-2A_{31} & -A_{32} & x & A_{34} \\
-3A_{41} & -2A_{42} & -A_{43} & x \\
\end{bmatrix}$$
I'm not sure if this is the correct way to approach the problem, as I'm having serious difficulty finding the eigenvalues from here. Any hints or help would be much appreciated.
 A: That's probably not the best approach. If $\lambda$ is an eigenvalue for $T_B$, then there exists $A$ with $$BA-AB=\lambda A.$$ That is, 
$$
\begin{bmatrix}
0&-A_{12}&-2A_{13}&-3A_{14}\\ 
A_{21}&0&-A_{23}&-2A_{24}\\ 
2A_{31}&A_{32}&0&-A_{34}\\ 
3A_{41}&2A_{42}&A_{43}&0\\ 
 \end{bmatrix}=\begin{bmatrix}
\lambda\,A_{11}&\lambda\,A_{12}&\lambda\,A_{13}&\lambda\,A_{14}\\ 
\lambda\,A_{21}&\lambda\,A_{22}&\lambda\,A_{23}&\lambda\,A_{24}\\ 
\lambda\,A_{31}&\lambda\,A_{32}&\lambda\,A_{33}&\lambda\,A_{34}\\ 
\lambda\,A_{41}&\lambda\,A_{42}&\lambda\,A_{43}&\lambda\,A_{44}\\ 
  \end{bmatrix}
$$
If $\lambda=0$, we get that $A$ is diagonal. 
If $\lambda\ne0$, then $A_{11}=A_{22}=A_{33}=A_{44}=0$. If $A_{12}\ne0$, then $\lambda=-1$. If $A_{21}\ne0$, then $\lambda=1$. If $A_{13}\ne0$, then $\lambda=-2$, and $\lambda=2$ if $A_{31}\ne0$. In a similar way we obtain $\lambda=\pm3$. So the eigenvalues are 
$$
\{-3,-2,-1,0,1,2,3\}.
$$
For the dimensions:


*

*$\lambda=0$: the eigenvectors are the diagonal matrices, so the dimension of the eigenspace is $4$.

*$\lambda=1$: this forces all coefficients to be equal to zero with the exception of $A_{21}$, $A_{32}$, and $A_{43}$. So the dimension of the eigenspace is $3$. 
-$\lambda=2$: here only $A_{31}$ and $A_{42}$ can be nonzero, so the eigenspace has dimension $2$. 


*

*$\lambda=3$: on $A_{41}$ can be nonzero, so the dimension of the eigenspace is $1$. 

*For the negative eigenvalues the logic is similar. 
In summary, the eigenvalues are $\lambda\in\{-3,-2,-1,0,1,2,3\}$ and the geometric multiplicity of $\lambda$ is $4-|\lambda|$. 
