Determinant of $T_A : S \rightarrow S \text{ where } T_A(X)=AXA^{T}$ for $S$ skew-symmetric and $A$ nonsingular Let $S$ denote the vector space of real $n \times n$ skew-symmetric
matrices. For a nonsingular matrix $A$, compute the determinant of the linear map
$T_A : S \rightarrow S \text{ where } T_A(X)=AXA^{T}$
No clue...
 A: With vectorization, we can identify $T_A$ as the restriction the operator $A \otimes A$ over $\Bbb R^{n} \otimes \Bbb R^n$ to the subspace $\Bbb R^n \wedge \Bbb R^n$. By the properties of the antisymmetric tensor product, if the eigenvalues of $A$ are $\lambda_1,\lambda_2,\dots,\lambda_n$, then the eigenvalues of $T_A$ will be of the form $\lambda_i\lambda_j$, where $i < j$.  It follows that the determinant will be
$$
\det(T_A) = \prod_{i<j} \lambda_i \lambda_j = \prod_i \lambda_i^{n-1} = \det(A)^{n-1}
$$
I am not aware of a more elementary proof of this result.

Idea for eigenvalue-free proof: Note that the matrices $E_{ij} = e_ie_j^T - e_{j}e_i^T := e_i \wedge e_j$ form an orthonormal basis of $S$ relative to a normalized Frobenius inner product.  As such, letting $[n]$ denote $[n] = \{1,\dots,n\}$ and letting $\mathcal S_{n \times n}$ denote the set of all permutations $\sigma:[n]^2 \to [n]^2$, we can write the determinant as
$$
\sum_{\sigma \in \mathcal S_{n \times n}}\sum_{(i,j) \in [n]^2}\langle T_A(E_{i,j}), E_{\sigma(i,j)}\rangle = \\
\sum_{(\sigma_1,\sigma_2) \in \mathcal S_{n \times n}}\sum_{(i,j) \in [n]^2}\langle T_A(E_{i,j}), E_{\sigma_1(i),\sigma_2(j)}\rangle = \\
\sum_{(\sigma_1,\sigma_2) \in \mathcal S_{n \times n}}\sum_{(i,j) \in [n]^2}\langle (Ae_i) \wedge (Ae_j), e_{\sigma_1(i)}\wedge e_{\sigma_2(j)}\rangle =\\
\sum_{(\sigma_1,\sigma_2) \in \mathcal S_{n \times n}}\sum_{(i,j) \in [n]^2}\det \pmatrix{\langle (Ae_i), \sigma_1(i)\rangle &  \langle (Ae_i), \sigma_2(j)\rangle\\
\langle (Ae_j), \sigma_1(i)\rangle &  \langle (Ae_j), \sigma_2(j)\rangle}
=\\
\sum_{(\sigma_1,\sigma_2) \in \mathcal S_{n \times n}}\sum_{(i,j) \in [n]^2}\det \pmatrix{A_{\sigma_1(i),i} &  A_{\sigma_2(j),i}\\
A_{\sigma_1(i),j} &  A_{\sigma_2(j),j}}
$$
Perhaps someone with more patience could take it from there.
A: I believe the following argument is valid.
Lets consider a base at $S$. If $E_{i,j}$ is the square matrix $n\times n$ given by $0$ at the coordinates different of $(i,j)$ and equal to 1 at the coordinate $(i,j)$ then the set $B$ of the matrix $b_{i,j}=E_{i,j}-E_{j,i}$ where $1 \leq i<j\leq n$, form a basis of $S$.
Lets suppose $A$ is a diagonal matrix. Lets write $A = [\lambda_1,\ldots,\lambda_n]$. We have $AE_{i,j} = \lambda_iE_{i,j}$. On the other hand $E_{i,j}A^T = \lambda_j E_{i,j}$. Therefore $AE_{i,j}A^T = \lambda_i\lambda_j E_{i,j}$. So, $T_Ab_{i,j} =\lambda_i \lambda_jb_{i,j}$.
The elements of the basis $B$ are eigenvalue of $T_A$. The determinant of $T_A$ is the product of the eigenvalues $\lambda_{i}\lambda_j$ with $1\leq i<j\leq n$. Therefore we have:
$$\det(T_A) = \prod_{1\leq i<j\leq n}\lambda_i\lambda_j.$$
Notice that $\lambda_1$ appears $n-1$ times on the product above because it appears on the products $\lambda_1\lambda_2,\ldots,\lambda_1,\lambda_n$. The same happens to $\lambda_2, \lambda_3$ etc. So we have:
$$\det T_A = (\lambda_1\cdots \lambda_n)^{n-1}.$$
Notice that $\det A  = \lambda_1\cdots \lambda_n$ and therefore $\det T_A = (\det A)^{n-1}$.
So the result is true in the case that $A$ is a diagonal matrix. If $A$ is diagonalizable then there is an invertible matrix $B$ and a diagonal matrix $C$ such that $A=B^{-1}CB$. Therefore $T_A = T_B \circ T_C \circ T_{B^{-1}}$ and $\det(T_A)=\det(T_C)$ since $T_B \circ T_{B^{-1}}=Id$. From $\det A =\det C$ we conclude $$\det T_A = \det T_C = (\det C)^{n-1} = (\det A)^{n-1}.$$
So, the result holds for diagonalizable matrices.
Now we use that the space of diagonalizable matrices is dense in the space of all matrices $n\times n$ and we obtain the result for a general matrix $A$ by continuity.
A: The answer is $\det T_A=\det A^{n-1}$. Here is a proof on the low road:


*

*Suppose $A$ is a diagonal matrix. Since $T_A(E_{ij}-E_{ji})=a_{ii}a_{jj}(E_{ij}-E_{ji})$ for each basis "vector" $E_{ij}-E_{ji}$ of $S$, we see that $\det T_A=\det A^{n-1}$ in this case.

*Next, suppose $A$ is real orthogonal. Then $T_A$ is an isometry and $\det T_A=\pm1$. Since $SO(n,\mathbb R)$ is connected and $T_A$ is continuous in $A$, we must have
$$
\det T_A=
\begin{cases}
\det T_I &\text{ when }\det A=1,\\
\det T_{\operatorname{diag}(1,\ldots,1,-1)} &\text{ when }\det A=-1.
\end{cases}
$$
It follows from (1) that $\det T_A=\det A^{n-1}$.

*Finally, for a generic matrix $A$, let $A=USV$ be a singular value decomposition. Then $T_A=T_UT_ST_V$. It follow from $(1)$ and $(2)$ that $\det T_A=\det A^{n-1}$.

