Question

Express the following matrix as the sum of a symmetric and skew-symmetric matrix and verify your result:

\[\begin{bmatrix}3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2\end{bmatrix}\] 

 

Answer 1

\[Here, \] 

\[ A = \begin{bmatrix}3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2\end{bmatrix}\] 
\[ \Rightarrow A^T = \begin{bmatrix}3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2\end{bmatrix}\] 
\[\text{Let X} = \frac{1}{2}\left( A + A^T \right) = \frac{1}{2}\left( \begin{bmatrix}3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2\end{bmatrix} + \begin{bmatrix}3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2\end{bmatrix} \right) = \begin{bmatrix}3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2\end{bmatrix}\] 
\[ X^T = \begin{bmatrix}3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2\end{bmatrix}^T = \begin{bmatrix}3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2\end{bmatrix} = X\] 
\[\text{Let Y} = \frac{1}{2}\left( A - A^T \right) = \frac{1}{2}\left( \begin{bmatrix}3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2\end{bmatrix} - \begin{bmatrix}3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2\end{bmatrix} \right) = \begin{bmatrix}0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0\end{bmatrix} \] 

\[ Y^T = \begin{bmatrix}0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0\end{bmatrix}^T = \begin{bmatrix}0 & \frac{5}{2} & \frac{3}{2} \\ \frac{- 5}{2} & 0 & 3 \\ \frac{- 3}{2} & - 3 & 0\end{bmatrix} = - \begin{bmatrix}0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0\end{bmatrix} = - Y\] 

\[ \text{Rightarrow X is a symmetric matrix and Y is a skew - symmetric matrix .} \] 

\[Now, \] 

\[ X + Y = \begin{bmatrix}3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2\end{bmatrix} + \begin{bmatrix}0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0\end{bmatrix} = \begin{bmatrix}3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2\end{bmatrix} = A\]