Dimension of Symmetric Matrix Space

hbghlyj

Let \(F\) be a field and \(n\) be a positive integer. Fix an \(n \times n\) matrix \(S\) over \(F\) that is invertible and symmetric. Writing \(A^t\) for the transpose of a matrix, we let
\[
V:=\left\{n \times n \text{ matrices } A \text { over } F: A^t=S A S^{-1}\right\}
\]
Note that \(V\) is a vector space. Find the dimension of \(V\) in terms of \(n\).

hbghlyj

The condition $A^t = S A S^{-1}$ defines the space $V$ of matrices that are self-adjoint with respect to the symmetric bilinear form induced by $S$. The dimension of this space is the same as the dimension of the space of symmetric matrices, which is $\frac{n(n+1)}{2}$.
To arrive at this, consider the linear map $\sigma: M_n(F) \to M_n(F)$ defined by $\sigma(A) = S^{-1} A^t S$. The space $V$ is the eigenspace of $\sigma$ corresponding to eigenvalue 1. Since $\sigma$ is an involution ($\sigma^2 = I$), the dimension of $V$ is $\frac{n^2 + \operatorname{tr}(\sigma)}{2}$, where $\operatorname{tr}(\sigma)$ is the trace of $\sigma$ as a linear map on the $n^2$-dimensional space $M_n(F)$.

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The trace $\operatorname{tr}(\sigma)$ is the trace of the linear map $\sigma: M_n(F) \to M_n(F)$ defined by $\sigma(A) = S^{-1} A^t S$.
To compute it, use the standard basis $\{E_{kl} \mid 1 \leq k,l \leq n\}$ for $M_n(F)$, where $E_{kl}$ is the matrix with 1 in position $(k,l)$ and 0 elsewhere. The trace is then $\sum_{k=1}^n \sum_{l=1}^n [\sigma(E_{kl})]_{kl}$, where $[\cdot]_{kl}$ denotes the $(k,l)$-entry of the matrix.
Now, $\sigma(E_{kl}) = S^{-1} E_{lk} S$, and the $(k,l)$-entry of this matrix is $(S^{-1})_{kl} S_{kl}$.
Thus, $\operatorname{tr}(\sigma) = \sum_{k=1}^n \sum_{l=1}^n (S^{-1})_{kl} S_{kl}$.
This sum is the Frobenius inner product $\langle S^{-1}, S \rangle = \operatorname{tr}((S^{-1})^t S)$. Since $S$ is symmetric and invertible, $S^{-1}$ is also symmetric, so $(S^{-1})^t = S^{-1}$ and $\operatorname{tr}((S^{-1})^t S) = \operatorname{tr}(S^{-1} S) = \operatorname{tr}(I_n) = n$.

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Using the standard basis $\{E_{ij}\}$ for $M_n(F)$, the trace $\operatorname{tr}(\sigma) = n$, leading to the dimension $\frac{n^2 + n}{2} = \frac{n(n+1)}{2}$.