Symmetric and skew-symmetric matrices

Question

Is the product of a symmetric and skew-symmetric matrix symmetric? does this answer remain accurate if the skew-symmetric matrix is replaced by a non-symmetric matrix? If not, what happens then?

Answer 1

No, the product of a symmetric matrix and a skew-symmetric matrix is not symmetric, just as the product of two symmetric matrices isn't symmetric either.

There special cases when squaring a matrix, such as the square of a symmetric matrix is also symmetric, and the square of a skew-symmetric matrix is also symmetric.

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Here is a overview of the structure of matrix multiplication

$$\begin{array}{cc} \text{operation} & \text{result}\\ \hline {\rm dense}*{\rm dense} & ={\rm dense}\\ {\rm symm}*{\rm dense} & ={\rm dense}\\ {\rm symm}*{\rm symm} & ={\rm dense}\\ {\rm symm}*{\rm skew} & ={\rm dense}\\ {\rm symm}*{\rm diag} & ={\rm dense}\\ {\rm skew}*{\rm dense} & ={\rm dense}\\ {\rm skew}*{\rm symm} & ={\rm dense}\\ {\rm skew}*{\rm skew} & ={\rm dense}\\ {\rm skew}*{\rm diag} & ={\rm dense}\\ {\rm diag}*{\rm dense} & ={\rm dense}\\ {\rm diag}*{\rm symm} & ={\rm dense}\\ {\rm diag}*{\rm skew} & ={\rm dense}\\ {\rm diag}*{\rm diag} & ={\rm diag}\\ {\rm dense}^{2} & ={\rm dense}\\ {\rm symm}^{2} & ={\rm symm}\\ {\rm skew}^{2} & ={\rm symm}\\ {\rm diag}^{2} & ={\rm diag} \end{array}$$

Answer 2

No product of a symmetric and skew-symmetric matrix is not symmetric. Let's take example. A is a symmetric matrix $$ A = \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} $$ $$ B = \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} $$ A is symmetric matrix and B is skew-symmetric matric. Multiplication of AB $$ AB = \begin{bmatrix} -1 & 1\\ -1 & 1 \end{bmatrix} $$ Transpose of AB $$ AB^T = \begin{bmatrix} -1 & -1\\ 1 & 1 \end{bmatrix} $$ As we can see AB is neither a symmetric and neither a skew symmetric. Soo product of symmetric and skew symmetric is not symmetric.

If you change skew symmetric matric with some other matrix then the product matrix will be symmetric or not will depend from matrix to matrix. Like $$ A = \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} $$ $$ B = \begin{bmatrix} x & y\\ x & y \end{bmatrix} $$ Let A be a symmetric matrix and B be the any normal matrix. Then $$AB = \begin{bmatrix} 2x & 2y\\ 2x & 2y \end{bmatrix} $$ If we take x = 1 and y =1 then AB will be the symmetric matrix $$(AB)^{'}= \begin{bmatrix} 2 & 2\\ 2 & 2 \end{bmatrix} $$ And if we take the x = 1 and y = 2 then AB will be non-symmetric matrix $$(AB)^{'}= \begin{bmatrix} 2 & 4\\ 2 & 4 \end{bmatrix} $$$$(AB)^{'T}= \begin{bmatrix} 2 & 2\\ 4 & 4 \end{bmatrix} $$