Let $A \in \mathbb{C}^{4 \times 4}$ be a diagonal matrix with main diagonal entries $1,2,3,4$. Define $T_A : \mathbb{C}^{4 \times 4} \rightarrow \mathbb{C}^{4 \times 4}$ by $T_A(B) = AB-BA$.
I am studying for a test and trying to answer some questions about $T_A$ but the answers I have come up with are very computational and I was wondering if there is a faster way to attack this problem then the method I will describe.
Part 1) What are $\dim(\ker(T_A))$ and $\dim(Image(T_A))$
I was thinking there might be some slick way to use the rank nullity theorem without actually having to do computation but so far my only answer involves calculating the matrix expressions:
$ AB = \left( \begin{array}{ccc} b_{11} & \ldots & b_{14} \\ 2b_{21} & \ldots & 2b_{24} \\ 3b_{31} & \ldots & 3b_{34} \\ 4b_{41} & \ldots & 4b_{44} \end{array} \right), BA = \left( \begin{array}{ccc} b_{11} & 2b_{12} & 3b_{13} & 4b_{14} \\ b_{11} & 2b_{12} & 3b_{13} & 4b_{14} \\ b_{11} & 2b_{12} & 3b_{13} & 4b_{14} \\ b_{11} & 2b_{12} & 3b_{13} & 4b_{14} \end{array} \right) $
To calculate the dimension of the kernel we just set $AB = BA$ and observe $2b_{21} = b{21}$ so $b_{21} =0$ and comparing the rest of the entries the same way we get $B \in \ker(T_A)$ when $B = \left( \begin{array}{ccc} b_{11} & 0 & 0 & 0 \\ 0 & 2b_{22} & 0 & 0 \\ 0 & 0 & 3b_{33} & 0 \\ 0 & 0 & 0 & 4b_{44} \end{array} \right) $
Therefore $\dim(\ker(T_A)) = 4$
Is there a faster way to do this?
Using $\dim(ker(T_A)) + \dim(Image(T_A)) = \dim(V)$
we can now see $4+ Image(T_A) = 16$ therefore $Image(T_A) = 12$
part 2) What are the eigenvalues of $T_A$
Using the equation $AB-BA = \lambda B$ and the calculations from part 1 implies
$ \left( \begin{array}{ccc} b_{11} & \ldots & b_{14} \\ 2b_{21} & \ldots & 2b_{24} \\ 3b_{31} & \ldots & 3b_{34} \\ 4b_{41} & \ldots & 4b_{44} \end{array} \right) - \left( \begin{array}{ccc} b_{11} & 2b_{12} & 3b_{13} & 4b_{14} \\ b_{21} & 2b_{22} & 3b_{23} & 4b_{24} \\ b_{31} & 2b_{32} & 3b_{33} & 4b_{34} \\ b_{41} & 2b_{42} & 3b_{43} & 4b_{44} \end{array} \right) = \left( \begin{array}{ccc} 0 & -b_{12} & -2b_{13} & -3b_{14} \\ b_{21} & 0 & -b_{23} & -2b_{24} \\ 2b_{31} & b_{32} & 0 & -b_{34} \\ 3b_{41} & 2b_{42} & b_{43} & 0 \end{array} \right) $
equating this with $\lambda B$ implies $B = \left( \begin{array}{ccc} 0 & 0 & 0 & 0 \\ b_{21} & 0 & 0 & 0 \\ 0 & b_{32} & 0 & 0 \\ 0 & 0 & b_{43} & 0 \end{array} \right) $
Which gives us the eigenvectors of $T_A$ with four repeated eigenvalues $\lambda =1$
We see the characteristic polynomial of $T_A = (x-1)^4$
Question 1) Is there a simple way to do any of part1) or part2) without resorting to all the matrix calculations. In particular I also want to compute the minimal polynomial of $T_A$ without having to compute powers of $(AB-BA-I)$.
Question2) Using the structure of the eigenvectors obtained in part 2) can we list a basis in which $T_A$ is diagonalizable?