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# Express the following matrix as the sum of a symmetric (B) and skew symmetric matrix (C) $\begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}$

$\begin{array}{1 1} B=\begin{bmatrix} 3 & \frac{-1}{2} & \frac{5}{2} \\ \frac{3}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2 \end{bmatrix} C = \begin{bmatrix} 1 & \frac{-5}{2} & \frac{-3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \\ \end{bmatrix} \\ B=\begin{bmatrix} 3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2 \end{bmatrix} C = \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ \frac{-5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \end{bmatrix} \\ B=\begin{bmatrix} 3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2 \end{bmatrix} C = \begin{bmatrix} 0 & \frac{-5}{2} & \frac{-3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \end{bmatrix}\\ B=\begin{bmatrix} 3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2 \end{bmatrix} C = \begin{bmatrix} 0 & \frac{-5}{3} & \frac{-3}{5} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \end{bmatrix}\end{array}$

Toolbox:
• Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix by $A=1/2(A+A’) +1/2(A-A’)$  Where $A+A’ \rightarrow$ symmetric matrix $A-A’ \rightarrow$ 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:
Given:
$A=\begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}$
Transpose can be obtained by changing the rows and column.
$A'=\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}$
Let $\frac{1}{2}$(A+A') = B
(A+A')=$\begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} +\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}=\begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix}$
$\frac{1}{2}(A+A')=\frac{1}{2}\begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix}=\begin{bmatrix} 3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2 \end{bmatrix}$
$\frac{1}{2}(A+A')$=B$\rightarrow$ Symmetric matrix
Step2:
Let $\frac{1}{2}(A-A')$ = C
(A-A')=$\begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} +(-1)\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}$
$\qquad=\begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} +\begin{bmatrix} -3 & -3 & 1 \\ 2 & 2 & - 1 \\ 4 & 5 & -2 \end{bmatrix}$
$\qquad=\begin{bmatrix} 0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0 \end{bmatrix}$
$\frac{1}{2}(A-A')=\begin{bmatrix} 0 & \frac{-5}{2} & \frac{-3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \end{bmatrix}$
$\frac{1}{2}(A-A')$=C$\rightarrow$Skew symmetric matrix.
Step3:
$\Rightarrow B+C=\begin{bmatrix} 3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2 \end{bmatrix}+\begin{bmatrix} 0 & \frac{-5}{2} & \frac{-3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \end{bmatrix}$
$\qquad\;\;\;=\begin{bmatrix} 3 & \frac{-4}{2} & \frac{-8}{2} \\ \frac{6}{2} & -2 & -1 \\ \frac{-2}{2} & 1 & 2 \end{bmatrix}$
$\qquad\;\;\;=\begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}=A.$

edited Dec 24, 2013