Let \(R\) be a commutative ring and let \(A, B \in M_n(R)\) be \(n \times n\) matrices.
Show that if \(A\) and \(B\) are symmetric (respectively, skew-symmetric), then \(A + B\) is symmetric (respectively, skew-symmetric).
Show that if \(A\) and \(B\) are symmetric, then the product \(A B\) is symmetric if and only if \(A B = B A\).
Show that for any matrix \(B \in M_n(R)\), the matrices \(B B^t\) and \(B + B^t\) are symmetric, and \(B - B^t\) is skew-symmetric.
Solution
1. For \(X, Y \in M_n(R)\): \((X + Y)^t = X^t + Y^t\), \((-X)^t=-X^t\), \((X^t)^t = X\), and \((X Y)^t = Y^t X^t\).
The first three identities hold entrywise. For the last, commutativity of \(R\) gives \[((XY)^t)_{i, j} = \sum_k X_{j, k} Y_{k, i} = \sum_k (Y^t)_{i, k} (X^t)_{k, j} = (Y^t X^t)_{i, j}.\]
2. (a) Sums of symmetric matrices are symmetric, and sums of skew-symmetric matrices are skew-symmetric.
1. If \(A^t = -A\) and \(B^t = -B\), then \((A + B)^t = (-A) + (-B) = -(A + B)\).
3. (b) If \(A\) and \(B\) are symmetric, then \(AB\) is symmetric if and only if \(AB = BA\).
By step 1, \((A B)^t = B^t A^t = B A\), so \((AB)^t=AB\) if and only if \(BA=AB\).
4. (c) \(B B^t\) and \(B + B^t\) are symmetric and \(B - B^t\) is skew-symmetric.
By step 1, \[(B B^t)^t = (B^t)^t B^t = B B^t,\qquad (B + B^t)^t = B^t + B,\qquad (B - B^t)^t = B^t - B = -(B - B^t).\]