# Express the following matrix as the sum of a symmetric

Question:

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

Solution:

Here,

$A=\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]$

$\Rightarrow A^{T}=\left|\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right|$

Let $X=\frac{1}{2}\left(A+A^{T}\right)=\frac{1}{2}\left(\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]+\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]\right)=\left[\begin{array}{ccc}3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2\end{array}\right]$

$X^{T}=\left[\begin{array}{ccc}3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2\end{array}\right]^{T}=\left[\begin{array}{ccc}3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2\end{array}\right]=X$

Let $Y=\frac{1}{2}\left(A-A^{T}\right)=\frac{1}{2}\left(\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]\right)=\left[\begin{array}{ccc}0 & \frac{-5}{2} & \frac{-3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0\end{array}\right]$

$Y^{T}=\left[\begin{array}{ccc}0 & \frac{-5}{2} & \frac{-3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0\end{array}\right]^{T}=\left[\begin{array}{ccc}0 & \frac{5}{2} & \frac{3}{2} \\ \frac{-5}{2} & 0 & 3 \\ \frac{-3}{2} & -3 & 0\end{array}\right]=-\left[\begin{array}{ccc}0 & \frac{-5}{2} & \frac{-3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0\end{array}\right]=-Y$

$\Rightarrow X$ is a symmetric matrix and $Y$ is a skew $-$ symmetric matrix.

Now,

$X+Y=\left[\begin{array}{ccc}3 & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ \frac{-5}{2} & -2 & 2\end{array}\right]+\left[\begin{array}{ccc}0 & \frac{-5}{2} & \frac{-3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0\end{array}\right]=\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=A$

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