Compute $\det{T}$ where $T(X)=AX+XA$ 
Consider the linear transformation $T:V\to V$ given by $T(X) = AX + XA$, where 
  $$A = \begin{pmatrix}1&1&0\\0&2&0\\0&0&-1 \end{pmatrix}.$$
  Compute the determinant $\det T$.

I know there was a similar problem with a different $A$, but that was a diagonal matrix, which made the situation easier. I computed $XA + AX$ using an arbitrary $X$, but I'm not sure where to go from there.
 A: For $A \in M_3(\mathbb{R})$, set $T_A(X) = AX + XA$. Note that your $A$ is diagonalizable and so we can find an invertible $P$ such that $P^{-1}AP = \operatorname{diag}(1,2,-1)$ := D. Consider the map $S \colon V \rightarrow V$ given by $S(X) = P^{-1}XP$. Note that $S$ is invertible and $S^{-1}(X) = PXP^{-1}$. Now,
$$ (S^{-1} \circ T_A \circ S)(X) = S^{-1}(T(P^{-1}XP)) = S^{-1}(AP^{-1}XP + P^{-1}XPA) = P(AP^{-1}XP + P^{-1}XPA)P^{-1} = DX + XD = T_D(X). $$
Thus, $T_A$ is similar to $T_D$ and $\det(T_A) = \det(T_D)$ so you can reduce the problem to the case the matrix is diagonal.
A: Note that $T$ is a linear transform for square matrices, so it makes more sense to rewrite $X\in\mathbb{R}^{n\times n}$ as a long vector $\mathrm{vec}(X)\in\mathbb{R}^{n^2}$, and rewrite $T:\mathbb{R}^{n\times n}\to\mathbb{R}^{n\times n}$ as the matrix representation of $T:\mathbb{R}^{n^2}\to\mathbb{R}^{n^2}$. We need Kronecker product for the latter representation. Basically,
$$
T(X)=AXB\Longrightarrow T=B^\mathrm{T}\otimes A
$$
In this case, we have $T=I\otimes A+A^\mathrm{T}\otimes I\in\mathbb{R}^{9\times 9}$. From the rule of calculating Kronecker product, $T$ can be divided into $3\times 3$ blocks. Denote the blocks as $T_{ij}\in \mathbb{R}^{3\times 3}\ (1\le i,j\le 3)$. From $A$'s shape, we know $T_{13}=T_{23}=T_{31}=T_{32}=\mathbf{0}^{3\times 3}$. Further we have $T_{33}=A-I\Longrightarrow\det T_{33}=0$. So finally we get $\det T=\det ([T_{11}\ T_{12};\ T_{21}\ T_{22}])\cdot\det T_{33}=0$.
Note: if explicitly write down all the stuff it will be the same as @mvw's answer.
A: One approach here is to "guess" the eigenvectors:
If $u$ is an eigenvector of $A$ and $v$ an eigenvector of $A^T$, then the matrix $uv^T$ is an "eigenmatrix" of the transformation $T$.
Now that you know the eigenvectors, you can find the eigenvalues, and find that the determinant is the product of all eigenvalues.

A more concise approach is as follows: plug in 
$$
X= \pmatrix{1\\0\\0}\pmatrix{0&0&1} = \pmatrix {0&0&1\\0&0&0\\0&0&0}
$$
to find that $T(X)=0$, which means that $T$ is not invertible, which means that $\det(T)=0$.
A: If $A$ is diagonalizable and its eigendecomposition is $A = Q \Lambda Q^{-1}$, then left-multiplying and right-multiplying both sides of $T (X) = A X + X A$ by $Q^{-1}$ and $Q$ produces
$$Q^{-1} T (X) Q = Q^{-1} A X Q + Q^{-1} X A Q = \Lambda Q^{-1} X Q + Q^{-1} X Q \Lambda = \Lambda Y + Y \Lambda$$
where $Y = Q^{-1} X Q$. Let $\tilde{T} (Y) := \Lambda Y + Y \Lambda$. Hence,
$$\det (T(X)) = \det (\Lambda Y + Y \Lambda) = \det (\tilde{T} (Y))$$
The image of the eigenmatrix $\mathrm{e}_i \mathrm{e}_j^T$ is
$$\tilde{T} (\mathrm{e}_i \mathrm{e}_j^T) = \Lambda \mathrm{e}_i \mathrm{e}_j^T + \mathrm{e}_i \mathrm{e}_j^T \Lambda = (\lambda_i \mathrm{e}_i) \mathrm{e}_j^T + \mathrm{e}_i (\lambda_j \mathrm{e}_j)^T = (\lambda_i + \lambda_j) \, \mathrm{e}_i \mathrm{e}_j^T$$
The eigenvalue associated with the eigenmatrix $\mathrm{e}_i \mathrm{e}_j^T$ is $\lambda_i + \lambda_j$. Thus, if we can find a pair of eigenvalues of $A$ such that $\lambda_i + \lambda_j = 0$, we can conclude that $\tilde{T}$ has a zero eigenvalue and that 
$$\det (\tilde{T}) = \det (T) = 0$$

Example:
The following matrix
$$A = \begin{bmatrix} 1 & 1 & 0\\ 0 & 2 & 0\\ 0 & 0 & -1\end{bmatrix}.$$
is diagonalizable and its spectrum is $\{2,1,-1\}$. Since $-1+1=0$, we conclude that $\det (T) = 0$
