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    For symmetric $A,B$, the product $AB$ is symmetric iff $AB=BA$; $BB^t$ and $B+B^t$ are symmetric

    P-HGQ8T

    Area
    Algebra
    Topics
    Bilinear Forms, Matrices
    Status
    draft

    Let \(A, B \in M_n(R)\) be \(n \times n\) matrices over a commutative ring \(R\). (1) Prove that if \(A\) and \(B\) are symmetric matrices (\(A^t = A, B^t = B\)), then the product \(AB\) is symmetric if and only if \(A\) and \(B\) commute (\(AB = BA\)). (2) For an arbitrary matrix \(B \in M_n(R)\), prove that \(B B^t\) and \(B + B^t\) are symmetric, and \(B - B^t\) is skew-symmetric.

    Solution

    Assume first that \(A^t=A\) and \(B^t=B\). Then \[ (AB)^t=B^tA^t=BA. \] Therefore \[ AB\text{ is symmetric} \iff (AB)^t=AB \iff BA=AB. \] This proves part (1).

    For arbitrary \(B\), \[ (BB^t)^t=(B^t)^tB^t=BB^t, \] so \(BB^t\) is symmetric. Also \[ (B+B^t)^t=B^t+B=B+B^t, \] while \[ (B-B^t)^t=B^t-B=-(B-B^t). \] Thus \(B+B^t\) is symmetric and \(B-B^t\) is skew-symmetric.

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