Question
Express the matrix $A =\left[\begin{array}{rrr}4 & 2 & -1 \\ 3 & 5 & 7 \\ 1 & -2 & 1\end{array}\right]$ as the sum of a symmetric and a skew-symmetric matrix.

Answer


$\begin{array}{l}\text { Given, } A=\left[\begin{array}{ccc}
4 & 2 & -1 \\
3 & 5 & 7 \\
1 & -2 & 1
\end{array}\right] \text { Then } A^{T}=\left[\begin{array}{ccc}
4 & 3 & 1 \\
2 & 5 & -2 \\
-1 & 7 & 1
\end{array}\right] \\
X=\frac{1}{2}\left(A+A^{T}\right) \\
=\frac{1}{2}\left(\left[\begin{array}{ccc}
4 & 2 & -1 \\
3 & 5 & 7 \\
1 & -2 & 1
\end{array}\right]+\left[\begin{array}{ccc}
4 & 3 & 1 \\
2 & 5 & -2 \\
-1 & 7 & 1
\end{array}\right]\right) \\
=\frac{1}{2}\left[\begin{array}{ccc}
4+4 & 2+3 & -1+1 \\
3+2 & 5+5 & 7-2 \\
1-1 & -2+7 & 1+1
\end{array}\right] \\
=\frac{1}{2}\left[\begin{array}{ccc}
8 & 5 & 0 \\
5 & 10 & 5 \\
0 & 5 & 2
\end{array}\right] \\
X=\left[\begin{array}{ccc}
4 & \frac{5}{2} & 0 \\
\frac{5}{2} & 5 & \frac{5}{2} \\
0 & \frac{5}{2} & 1
\end{array}\right] \\
X^{T}=\left[\begin{array}{lll}
4 & \frac{5}{2} & 0 \\
\frac{5}{2} & 5 & \frac{5}{2} \\
0 & \frac{5}{2} & 1
\end{array}\right]^{T}=\left[\begin{array}{lll}
4 & \frac{5}{2} & 0 \\
\frac{5}{2} & 5 & \frac{5}{2} \\
0 & \frac{5}{2} & 1
\end{array}\right]=X
\end{array}
$
$\therefore X$ is a symmetric matrix.
$
\begin{array}{l}
Y=\frac{1}{2}\left(A-A^{T}\right) \\
=\frac{1}{2}\left(\left[\begin{array}{ccc}
4 & 2 & -1 \\
3 & 5 & 7 \\
1 & -2 & 1
\end{array}\right]-\left[\begin{array}{ccc}
4 & 3 & 1 \\
2 & 5 & -2 \\
-1 & 7 & 1
\end{array}\right]\right) \\
=\frac{1}{2}\left[\begin{array}{ccc}
4-4 & 2-3 & -1-1 \\
3-2 & 5-5 & 7+2 \\
1+1 & -2-7 & 1-1
\end{array}\right] \\
=\frac{1}{2}\left[\begin{array}{ccc}
0 & -1 & -2 \\
1 & 0 & 9 \\
2 & -9 & 0
\end{array}\right] \\
Y=\left[\begin{array}{ccc}
0 & -\frac{1}{2} & -1 \\
\frac{1}{2} & 0 & \frac{9}{2} \\
1 & -\frac{9}{2} & 0
\end{array}\right] \\
Y^{T}=\left[\begin{array}{ccc}
0 & -\frac{1}{2} & -1 \\
\frac{1}{2} & 0 & \frac{9}{2} \\
1 & -\frac{9}{2} & 0
\end{array}\right]^{T}=\left[\begin{array}{ccc}
0 & -\frac{1}{2} & -1 \\
\frac{1}{2} & 0 & \frac{9}{2} \\
1 & -\frac{9}{2} & 0
\end{array}\right]=Y
\end{array}
$
$\Rightarrow Y$ is a skew-symmetric matrix.
Now,
$X+Y=\left[\begin{array}{ccc}4 & \frac{5}{2} & 0 \\ \frac{5}{2} & 5 & \frac{5}{2} \\ 0 & \frac{5}{2} & 1\end{array}\right]+\left[\begin{array}{ccc}0 & -\frac{1}{2} & -1 \\ \frac{1}{2} & 0 & \frac{9}{2} \\ 1 & -\frac{9}{2} & 0\end{array}\right]$
$
\begin{array}{l}
=\left[\begin{array}{ccc}
4+0 & \frac{5}{2}-\frac{1}{2} & 0-1 \\
\frac{5}{2}+\frac{1}{2} & 5+0 & \frac{5}{2}+\frac{9}{2} \\
0+1 & \frac{5}{2}-\frac{9}{2} & 1+0
\end{array}\right] \\
=\left[\begin{array}{ccc}
4 & 2 & -1 \\
3 & 5 & 7 \\
1 & -2 & 1
\end{array}\right]=A
\end{array}
$
Hence, $X + Y = A$.
Thus matrix A is expressed as the sum of symmetric and skew-symmetric matrices.

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