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From Berkeley Problems in Mathematics, Summer 1981. 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 $$T_{A}:S\rightarrow S, T_{A}(X)=AXA^{t}$$

My main difficulty is I do not know how to associate $S$ and $T_{A}$ together. $S$ have $\frac{n^{2}-n}{2}$ dimension, and $T_{A}(X)_{ij}=\sum\sum\sum a_{ik}x_{km}a_{jm}$. Thus the entry for the linear transformation $T_{A}(X)_{ij}=\sum \sum a_{ik}a_{jm}$ looks nothing but the product of $\sum a_{ik}\sum a_{jm}$. I do not know how to solve it in better methods (find the eigenvalue was my first thought).

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In the first expression for $T_A(X)_{ij}$, I think you intended the first $j$ to be a $k$? Then you give a second expression for $T_A(X)_{ij}$ -- how is this compatible with the first, and where does it come from? –  joriki Jul 28 '12 at 17:43
    
There is a typo indeed. The second expression means considering $T_{A}$ as a map from the $\frac{n^{2}-n}{2}$ dimensional space $S$ to itself, and if we write down the matrix the coefficient at $\{i,j\}$ is the expression I give. –  Bombyx mori Jul 28 '12 at 17:47
    
I don't understand that. Either $i$ and $j$ run from $1$ to $n$, then this isn't a coefficient of an $(n^2-n)/2\times(n^2-n)/2$ matrix; or they run from $1$ to $(n^2-n)/2$; then $a_{ik}$ and $a_{jm}$ make no sense. –  joriki Jul 28 '12 at 17:51
    
That is true indeed, let me think about this. –  Bombyx mori Jul 28 '12 at 17:53

2 Answers 2

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Suppose $A$ is diagonalizable, then there is a basis of eigenvectors $v_n$ corresponding to the eigenvalues $\lambda_n$. It is straightforward to verify that $V_{ij} = v_i v_j^T - v_j v_i^T$, for all $i>j$, is a basis for $S$.

Now consider $T_A(V_{ij}) = A V_{ij} A^T = A(v_i v_j^T - v_j v_i^T)A^T = \lambda_i (v_i v_j^T - v_j v_i^T) \lambda_j = \lambda_i \lambda_j V_{ij}$. Hence, $T_A$ is diagonal in this basis, and we have $\det T_A = \prod_{i>j} \lambda_i \lambda_j = \prod_k \lambda_k^{n-1} = (\det A)^{n-1}$.

Now we note that the diagonalizable matrices are dense, and the determinant and eigenvalues are continuous functions of the matrix $A$. Hence the result is true for all matrices $A$.

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First consider the case where $A$ is a diagonal matrix with entries $c_1, c_2 \ldots, c_n$. Then it is easy to check that the matrix with 1 in the $(i,j)$ position and -1 in the $(j,i)$ position, and 0 elsewhere, is an eigenvector with eigenvalue $c_i c_j$. Since there are $n(n-1)/2 = \dim S$ such eigenvectors, these are all the eigenvectors, so the determinant of $T_A$ is $\prod_{1 \le i<j \le n} c_i c_j = (\det A)^{n-1}$.

Next, let's tackle the case where $A$ is diagonalizable. If we replace $A$ by a conjugate $PAP^{-1}$, the relationship between $T_A$ and $T_{PAP^{-1}}$ is $$T_{PAP^{-1}}(PXP^t) = P T_A(X) P^t$$ Hence $T_{PAP^{-1}}$ is conjugate to $T_A$ under the linear transformation $X \mapsto PXP^t$ of $S$, so they have the same determinant. So if $P$ is a matrix such that $PAP^{-1}$ is diagonal, then using the previous case,$$\det T_A = \det T_{PAP^{-1}} = (\det PAP^{-1})^{n-1} = (\det A)^{n-1}$$

The general case should follow by a continuity argument, or by generalizing the first paragraph to Jordan canonical form matrices, instead of just diagonal matrices.

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Thanks, I shall try to generalize. –  Bombyx mori Jul 28 '12 at 17:48

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