Chapter 2 - Matrices and Systems of Linear Equations - 2.2 Matrix Algebra - Problems - Page 137: 32

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We know $A$ is an $n × n$ matrix, then $B^T=[\frac{1}{2}(A+A^T)]^T=\frac{1}{2}(A^T+A)=B$ Thus, $B$ is a symmetric matrix. We obtain: $C^T=[\frac{1}{2}(A-A^T)]^T=\frac{1}{2}(A^T-A)=-C$ Hence, $C$ is a skew-symmetric matrix.