Express the following matrix as the sum of a symmetric and skew-symmetric matrix and verify your result: $[\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}]$ .

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Solution

$Here, A = [\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}] \Rightarrow A^{T} = [\begin{matrix} 3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2 \end{matrix}] Let X = \frac{1}{2} (A + A^{T}) = \frac{1}{2} ([\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}] + [\begin{matrix} 3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2 \end{matrix}]) = [\begin{matrix} 3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}] X^{T} = {[\begin{matrix} 3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}]}^{T} = [\begin{matrix} 3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}] = X Let Y = \frac{1}{2} (A - A^{T}) = \frac{1}{2} ([\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}] - [\begin{matrix} 3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2 \end{matrix}]) = [\begin{matrix} 0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0 \end{matrix}] Y^{T} = {[\begin{matrix} 0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0 \end{matrix}]}^{T} = [\begin{matrix} 0 & \frac{5}{2} & \frac{3}{2} \\ \frac{- 5}{2} & 0 & 3 \\ \frac{- 3}{2} & - 3 & 0 \end{matrix}] = - [\begin{matrix} 0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0 \end{matrix}] = - Y \Rightarrow X is a symmetric matrix and Y is a skew - symmetric matrix . Now, X + Y = [\begin{matrix} 3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}] + [\begin{matrix} 0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0 \end{matrix}] = [\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}] = A$

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