Given the linear map $T:V\rightarrow V$ such that $T(B)=AB+BA$ for a given A; what is the characteristic polynomial in terms of A? Given the linear map $T:V\rightarrow V$, where $V$ is the space of $2\times2$ complex matrices and such $T(B)=AB+BA$ for a given $A$; what is the characteristic polynomial in terms of the trace an and determinant of $A$?
We are told first to consider the case when $A$ is diagonalizable and I think once I have got this part I know how to extend it to general $A$. I just don't see how we can express the characteristic polynomial purely in terms of $A$.So far I have found the characteristic polynomial of a general $2\times2$ matrix C to be $\det(C)+Tr(C)^2t+Tr(C)t^2$ but this doesn't seem to help much; I just don't see how the diagonalizability of A helps us. Hints please!
 A: If $A$ is diagonalisable, then we have $A v_k = \lambda_k v_k$ for some basis $v_k$, and $u_k^* A = \lambda_k u_k^*$ for some dual basis $u_k^*$.
Check that $u_i v_j^*$ form a basis for the matrices.
Compute $T(u_i v_j^T)$, this gives eigenvalues & eigenvectors of $T$.
Now note that $\chi_T(s) = \prod_{i,j} (s - (\lambda_i+\lambda_j))$.
If you expand the latter, then the coefficients of $1,s,s^2,s^3$ can
all be written in terms of $\lambda_1+\lambda_2 = \operatorname{tr} A$ and
$\lambda_1 \lambda_2 = \det A$.
For a general $A$, we can find a sequence of diagonalisable $A_n$ such that
$A_n \to A$. Let $T_A$ denote the corresponding operator.
Since there is some continuous $\phi$ such that
$\chi_{A_n}(s)=\det(sI -T_{A_n}) = \phi(\operatorname{tr} {A_n}, \det {A_n}, s)$, we see that $\chi_{A}(s)=\phi(\operatorname{tr} {A}, \det {A}, s)$,
that is, the same formula holds as for the diagonalisable case.
A: The $2\times 2$ matrices here are the "vectors", $V$. The linear operator $T$ acts on this 4-dimensional vector space. 
When $A=\begin{bmatrix}\alpha&0\\0&\beta\end{bmatrix}$, we have
$$
(T-\lambda I)B=\begin{bmatrix}(2\alpha-\lambda)b_{11}&(\text{Tr}(A)-\lambda)b_{12}\\ (\text{Tr}(A)-\lambda)b_{21}&(2\beta-\lambda)b_{22}\end{bmatrix}.
$$
For $\lambda$ to be an eigenvalue of $T$, we need that some nonzero $B$ makes the whole matrix equal to zero. This can only happen when  $\lambda$ is $2\alpha$, $2\beta$, or $\text{Tr}(A)$; this last one with dimension $2$. As $T$ acts on a 4-dimensional space, these are all the eigenvalues. So the characteristic polynomial of $T$ is 
$$
p(t)=(t-2\alpha)(t-2\beta)(t-\text{Tr}(A))^2=(t^2-2\text{Tr}(A)\,t+4\det(A))(t-\text{Tr}(A))^2
$$
(since $\alpha+\beta=\text{Tr}(A)$ and $\alpha\beta=\det(A)$).
It is not hard to check that this extends well to the case where $A$ is diagonalizable: if $A=VDV^{-1}$, then 
$$
(T-\lambda I)B=AB+BA-\lambda B=v(DV^{-1}BV+V^{-1}BVD-\lambda V^{-1}BV)V^{-1},
$$
so one can apply the above with $D$ in place of $A$ and $V^{-1}BV$ in place of $B$. 
When $A$ is not diagonalizable, the only possibility is that the Jordan form of $A$ is 
$$
J=\begin{bmatrix}\alpha&1\\0&\alpha\end{bmatrix}.
$$
In this case, assuming first that $A=J$, 
$$
TB-\lambda B=\begin{bmatrix} (2\alpha-\lambda)b_{11}+b_{21}&(2\alpha-\lambda)b_{12}+b_{11}+b_{22} \\ (2\alpha-\lambda)b_{21}&(2\alpha-\lambda)b_{22}+b_{21}\end{bmatrix}.
$$
If $\lambda\ne2\alpha$, this can only be zero when $B=0$, so $\alpha$ is not an eigenvalue. If $\lambda=2\alpha$, one can check explicitly that $(T-2\alpha I)^2=0$, so that's the minimal polynomial:
$$
p(t)=(t-2\alpha)^2=(t-\text{Tr}(A))^2.
$$
The characteristic polynomial is then $p(t)=(t-\text{Tr}(A))^4$, which agrees exactly with the previous case:
$$
t^2-2\text{Tr}(A)\,t+4\det(A)=t^2-4\alpha\,t+4\alpha^2=(t-2\alpha)^2=(t-\text{Tr}(A))^2. 
$$
As in the other case, replacing $A$ with $VAV^{-1}$ will not change the characteristic polynomial (this needs to be done, it is not the usual case as these are not $2\times 2$ matrices acting on $\mathbb R^2$). So the characteristic polynomial of $T$ is (always!)
$$
p(t)=(t^2-2\text{Tr}(A)\,t+4\det(A))\,(t-\text{Tr}(A))^2.
$$
