Question

Express$\left[\begin{array}{ccc}6& -4& 5\\ 1& 4& -2\\ 7& 5& 9\end{array}\right]$ as a sum of a symmetric matrix and a skew-symmetric matrix.

Answer 1

Let $A=\left[\begin{array}{ccc}6& -4& 5\\ 1& 4& -2\\ 7& 5& 9\end{array}\right]$

Therefore,

$A^{\prime }=\left[\begin{array}{ccc}6& 1& 7\\ -4& 4& 5\\ 5& -2& 9\end{array}\right]$

Now,

$A+A^{\prime }=\left[\begin{array}{ccc}6& -4& 5\\ 1& 4& -2\\ 7& 5& 9\end{array}\right]+\left[\begin{array}{ccc}6& 1& 7\\ -4& 4& 5\\ 5& -2& 9\end{array}\right]=\left[\begin{array}{ccc}12& -3& 12\\ -3& 8& 3\\ 12& 3& 18\end{array}\right]$

$\Rightarrow {(A+A^{\prime })}^{\prime }=\left[\begin{array}{ccc}12& -3& 12\\ -3& 8& 3\\ 12& 3& 18\end{array}\right]=(A+A^{\prime })\Rightarrow$ Symmetric matrix

Similarly,

$A-A^{\prime }=\left[\begin{array}{ccc}6& -4& 5\\ 1& 4& -2\\ 7& 5& 9\end{array}\right]-\left[\begin{array}{ccc}6& 1& 7\\ -4& 4& 5\\ 5& -2& 9\end{array}\right]=\left[\begin{array}{ccc}0& -5& -2\\ 5& 0& -7\\ 2& 7& 0\end{array}\right]$

$\Rightarrow {(A-A^{\prime })}^{\prime }=\left[\begin{array}{ccc}0& 5& 2\\ -5& 0& 7\\ -2& -7& 0\end{array}\right]=-(A-A^{\prime })\Rightarrow$ Skew symmetric matrix

Now,

$\therefore A=\underset{\left(\text{Symmetric matrix}\right)}{\frac{1}{2}(A+A)}+\underset{\left(\text{Skew - symmetric matrix}\right)}{\frac{1}{2}(A-A^{\prime })}$