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# Express the following matrices as the sum of a symmetric and a skew symmetric matrix: $\quad \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Toolbox:
• Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix by $A=\frac{1}{2}(A+A') +\frac{1}{2}(A-A')$ Where A+A' --> symmetric matrix A-A' --> Skew symmetric matrix
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
Step1:
Let B=1/2(A+A') [Symmetric matrix]
Let (i)$A=\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$
$A'=\begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$
$\frac{1}{2}(A+A')=\frac{1}{2}\bigg(\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}+\begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} \bigg)$
$(A+A')=\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}+\begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3+3 & 5+1 \\1+ 5 & -1-1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix}$
$\;\;\;=1/2\begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} \rightarrow B$ symmetric matrix
Step2:
Let C=1/2(A-A') [Skew symmetric matrix]
$\frac{1}{2}(A-A')=\frac{1}{2}\bigg(\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}+(-)\begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} \bigg)$
$(A-A')=\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}+\begin{bmatrix} -3 & -1 \\- 5 & 1 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3-3 & 5-1 \\ 1-5 & -1+1 \end{bmatrix}$
$A'=\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}$
$\frac{1}{2}(A-A')=\begin{bmatrix}0 & 2\\-2 & 0\end{bmatrix}\rightarrow C$ skew symmetric matrix
Step3:
$B+C=\begin{bmatrix} 3 & 3 \\ 3 & -1\end{bmatrix} +\begin{bmatrix}0 & 2\\-2 & 0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}3+0 &3+ 2\\3-2 & -1+0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}3 & 5\\1 & -1\end{bmatrix}=A.$
Thus we get sum of a symmetric and skew symmetric matrix.

edited Mar 21, 2013