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# Express the following matrices as the sum of a symmetric and a skew symmetric matrix: $(iv) \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}$

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:
(iv)Let B=1/2(A+A') [Symmetric matrix]
Let $A=\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}$
$A'=\begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}$
$A+A'=\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}+\begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 1+1 & 5-1 \\ -1+5 & 2+2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 2 & 4 \\ 4 & 4\end{bmatrix}$
$\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}\rightarrow B$ -symmetric matrix
Step2:
Let C=1/2(A-A') [Skew symmetric matrix]
$A-A'=\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}+(-1)\begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}$
$A-A'=\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}+\begin{bmatrix} -1 & 1 \\- 5 & -2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 1-1 & 5+1 \\ -1-5 & 2-2 \end{bmatrix}=\begin{bmatrix}0 & 6\\-6 &0 \end{bmatrix}$
$\frac{1}{2}(A-A')=\frac{1}{2}\begin{bmatrix}0 & 6\\-6 & 0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}0 & 3\\-3 & 0\end{bmatrix}\rightarrow C$-skew symmetric matrix.
$B+C=\begin{bmatrix}1 & 2\\2 &2\end{bmatrix}+\begin{bmatrix}0 & 3\\-3 & 0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}1+0 & 2+3\\2-3 & 2+0\end{bmatrix}\Rightarrow \begin{bmatrix}1 & 5\\-1 & 2\end{bmatrix}=A$

edited Mar 21, 2013