I am trying to solve the following problem:
Let $A$ be a $2\times 2$ complex matrix. Show that the linear map $$B\longmapsto AB + BA$$ is a bijection if and only if $\det(A)$ and $\text{tr}(A)$ are nonzero.
My attempt: Let us call $T$ the linear map. The Jordan canonical form of $A$ can be of the form $\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$ or $\begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}$.
Suppose that $A$ is diagonalizable. Then, there exist vectors $v_1$ and $v_2$ in $\mathbb{C}^2$ such that $Av_1 = \lambda_1 v_1$, $Av_2 = \lambda_2 v_2$, and $\{v_1,v_2\}$ is a basis of $\mathbb{C}^2$. It follows that the set of the matrices $$X_1 = \begin{pmatrix} v_1 & 0 \end{pmatrix}, ~~X_2 = \begin{pmatrix} 0&v_1 \end{pmatrix},~~ X_3 = \begin{pmatrix} v_2 & 0 \end{pmatrix},~~X_4 = \begin{pmatrix} 0 & v_2 \end{pmatrix}$$ is a basis of $M_2(\mathbb{C})$.
Notice that $$T(X_1) = AX_1 + X_1 A = \begin{pmatrix} \lambda_1v_1 & 0 \end{pmatrix} + \begin{pmatrix} a_{11}v_1 & a_{12}v_1 \end{pmatrix} = \begin{pmatrix} (\lambda_1+a_{11})v_1 & a_{12}v_1 \end{pmatrix} = (\lambda_1+a_{11})X_1 + a_{12}X_2$$ and $$T(X_2) = AX_2 + X_2 A = \begin{pmatrix} 0 & \lambda_1v_1 \end{pmatrix} + \begin{pmatrix} a_{21}v_1 & a_{22}v_1 \end{pmatrix} = \begin{pmatrix} a_{21}v_1 & (\lambda_1+a_{22})v_1 \end{pmatrix} = a_{21}X_1 + (\lambda_1+a_{22})X_2.$$ Likewise, we get that $T(X_3) = (\lambda_2+a_{11})X_3 + a_{12}X_4$ and $T(X_4) = a_{21}X_3 + (\lambda_2+a_{22})X_4$. Thus, the representation of $T$ on the basis $\{X_1,X_2,X_3,X_4\}$ is given by the matrix:
$$
\begin{pmatrix}
\lambda_1+a_{11} & a_{21} & 0 & 0 \\
a_{12}& \lambda_1+a_{22} & 0 & 0 \\
0 & 0 & \lambda_2+a_{11} & a_{21} \\
0 & 0 & a_{12} & \lambda_2+a_{22} \\
\end{pmatrix}.
$$
Since this matrix is diagonal by blocks, we have that $$\det(T) = \det\begin{pmatrix}
\lambda_1+a_{11} & a_{21} \\
a_{12}& \lambda_1+a_{22}
\end{pmatrix} \det\begin{pmatrix}
\lambda_2+a_{11} & a_{21} \\
a_{12} & \lambda_2+a_{22}
\end{pmatrix} = (\lambda_1^2+\text{tr}(A)\lambda_1+\det(A))(\lambda_2^2+\text{tr}(A)\lambda_2+\det(A)).$$ Because $\lambda_1$ and $\lambda_2$ are roots of the polynomial $x^2-\text{tr}(A)x+\det(A)$, it follows that $$\det(T) = (2\text{tr}(A)\lambda_1)(2\text{tr}(A)\lambda_2) = 4\text{tr}(A)^2\det(A).$$ Hence, $\det(T)=0$ if and only if $\det(A)=0$ or $\text{tr}(A)=0$, which completes the proof of this case.
However, I cannot apply the same idea to prove the statement when $A$ is not diagonalizable, so I am stuck. Can you give me some advice to complete the exercise? Thanks in advance.
Edit: I just realized that for the case $J_A = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}$ we can use a similar idea. Indeed, there exists a basis $\{v_1,v_2\}$ of $\mathbb{C}^2$ such that $Av_1 = \lambda v_1$ and $Av_2 = v_1 + \lambda v_2$. Repeating the previous construction with this basis, we get the following representation for T:$$
\begin{pmatrix}
\lambda+a_{11} & a_{21} & 1 & 0 \\
a_{12}& \lambda+a_{22} & 0 & 1 \\
0 & 0 & \lambda+a_{11} & a_{21} \\
0 & 0 & a_{12} & \lambda+a_{22} \\
\end{pmatrix}.
$$
This matrix also has determinant equal to $4\text{tr}(A)^2\det(A)$, and so the problem is solved.
I am still looking for a shorter approach.
Best Answer
If $\det(A)=0$, there exist two nonzero vectors $u$ and $v$ such that $Au=0$ and $v^TA=0$. It follows that $T(uv^T)=0$.
If $\det(A)\ne0$ but $\operatorname{tr}(A)=0$, then $A$ has two nonzero eigenvalues $\lambda$ and $-\lambda$. Pick $x,y\ne0$ such that $Ax=\lambda x$ and $y^TA=-\lambda y^T$. Then $T(xy^T)=0$.
Therefore $T$ is singular when $\det(A)=0$ or $\operatorname{tr}(A)=0$.
Conversely, if $T$ is singular, there exists a nonzero matrix $B$ such that $T(B)=AB+BA=0$. Hence $AB=-BA$ and in turn, $A^2B=BA^2$. As $A^2=\operatorname{tr}(A)A-\det(A)I_2$ (Cayley-Hamilton theorem), the previous equality implies that $\operatorname{tr}(A)AB=\operatorname{tr}(A)BA$. So, we must have $\operatorname{tr}(A)=0$ or $AB=BA$. In the latter case, since we also have $AB=-BA$, we obtain $AB=0$. Since $B$ is nonzero, $A$ must be singular and $\det(A)=0$ in this case.