Let \(S\) be the vector space of real \(n\times n\) skew-symmetric matrices. For \(A\in GL_n(\mathbb R)\), compute the determinant of the linear map \[ T_A:S\to S, \qquad T_A(X)=AXA^T. \]
Solution
Let \[ V=\RR^n. \]
1. The map\[ \Phi:\bigwedge^2V\longrightarrow S \]defined on decomposable elements by\[ \Phi(u\wedge v) = uv^T-vu^T \]is a linear isomorphism.
For the standard basis \(e_1,\ldots,e_n\) of \(V\), \[ \Phi(e_i\wedge e_j) = E_{ij}-E_{ji} \] for \(i<j\). The vectors \[ e_i\wedge e_j, \qquad i<j, \] form a basis of \(\bigwedge^2V\), while the matrices \[ E_{ij}-E_{ji}, \qquad i<j, \] form a basis of \(S\). Thus \(\Phi\) sends a basis to a basis.
2. Under the isomorphism \(\Phi\), the map \(T_A\) corresponds to theexterior-square operator\[ \bigwedge^2A:\bigwedge^2V\longrightarrow\bigwedge^2V. \]
For \(u,v\in V\), \[ \begin{aligned} T_A(\Phi(u\wedge v)) &= A(uv^T-vu^T)A^T\\ &= (Au)(Av)^T-(Av)(Au)^T\\ &= \Phi(Au\wedge Av)\\ &= \Phi\bigl((\bigwedge^2A)(u\wedge v)\bigr). \end{aligned} \] Since decomposable wedges span \(\bigwedge^2V\), the intertwining identity holds on the whole space.
3. Consequently,\[ \det T_A = \det(\bigwedge^2A). \]
Step 2 says that \[ T_A = \Phi\circ(\bigwedge^2A)\circ\Phi^{-1}. \] Similar linear transformations have the same determinant.
4. Let\[ \lambda_1,\ldots,\lambda_n \]be the complex eigenvalues of \(A\), counted with algebraic multiplicity.Then\[ \det(\bigwedge^2A) = \prod_{1\leq i<j\leq n}\lambda_i\lambda_j. \]
Complexify the real vector space and the operator. The matrix of the complexified operator is still \(A\), so its determinant is unchanged.
Over \(\CC\), choose a basis \(f_1,\ldots,f_n\) of \(\CC^n\) in which \(A\) is upper triangular with diagonal entries \[ \lambda_1,\ldots,\lambda_n. \] In the induced wedge basis \[ f_i\wedge f_j, \qquad i<j, \] ordered lexicographically, the operator \(\bigwedge^2A\) is triangular, and its diagonal entry corresponding to \(f_i\wedge f_j\) is \[ \lambda_i\lambda_j. \] The determinant of a triangular matrix is the product of its diagonal entries, giving the displayed formula.
5. One has\[ \prod_{1\leq i<j\leq n}\lambda_i\lambda_j = \left( \prod_{i=1}^n\lambda_i \right)^{n-1}. \]
Fix \(i\). The eigenvalue \(\lambda_i\) appears once in the factor \(\lambda_i\lambda_j\) for every \(j\neq i\), hence exactly \(n-1\) times in the full product. Therefore the total product is \[ \prod_{i=1}^n\lambda_i^{\,n-1} = \left( \prod_{i=1}^n\lambda_i \right)^{n-1}. \]
6. Since\[ \prod_{i=1}^n\lambda_i=\det A, \]the determinant of \(T_A\) is\[ \boxed{ \det T_A=(\det A)^{n-1}. } \]
Combine steps <1>3–<1>5 and the standard identity that the product of the eigenvalues of a matrix, counted with algebraic multiplicity, equals its determinant.
7. Q.E.D.
Step 6 is the requested determinant.