# Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

Question:

Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i) $\left[\begin{array}{rr}3 & 5 \\ 1 & -1\end{array}\right]$

(ii) $\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$

(iii) $\left[\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]$

(iv) $\left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]$

Solution:

(i)

Let $A=\left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right]$, then $A^{\prime}=\left[\begin{array}{cc}3 & 1 \\ 5 & -1\end{array}\right]$

Now, $A+A^{\prime}=\left[\begin{array}{rr}3 & 5 \\ 1 & -1\end{array}\right]+\left[\begin{array}{rr}3 & 1 \\ 5 & -1\end{array}\right]=\left[\begin{array}{rr}6 & 6 \\ 6 & -2\end{array}\right]$

Let $P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rr}6 & 6 \\ 6 & -2\end{array}\right]=\left[\begin{array}{rr}3 & 3 \\ 3 & -1\end{array}\right]$

Now, $P^{\prime}=\left[\begin{array}{rr}3 & 3 \\ 3 & -1\end{array}\right]=P$

Thus, $P=\frac{1}{2}\left(A+A^{\prime}\right)$ is a symmetric matrix.

Now, $A-A^{\prime}=\left[\begin{array}{rr}3 & 5 \\ 1 & -1\end{array}\right]-\left[\begin{array}{rr}3 & 1 \\ 5 & -1\end{array}\right]=\left[\begin{array}{rr}0 & 4 \\ -4 & 0\end{array}\right]$

Let $Q=\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rr}0 & 4 \\ -4 & 0\end{array}\right]=\left[\begin{array}{rr}0 & 2 \\ -2 & 0\end{array}\right]$

Now, $Q^{\prime}=\left[\begin{array}{rr}0 & 2 \\ -2 & 0\end{array}\right]=-Q$

Thus, $Q=\frac{1}{2}\left(A-A^{\prime}\right)$ is a skew-symmetric matrix.

Representing $A$ as the sum of $P$ and $Q$ :

$P+Q=\left[\begin{array}{rr}3 & 3 \\ 3 & -1\end{array}\right]+\left[\begin{array}{rr}0 & 2 \\ -2 & 0\end{array}\right]=\left[\begin{array}{rr}3 & 5 \\ 1 & -1\end{array}\right]=A$

(ii)

Let $A=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$, then $A^{\prime}=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$

Now, $A+A^{\prime}=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]+\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]=\left[\begin{array}{rrr}12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6\end{array}\right]$

Let $P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rrr}12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6\end{array}\right]=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$

Now, $P^{\prime}=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]=P$

Thus, $P=\frac{1}{2}\left(A+A^{\prime}\right)$ is a symmetric matrix.

Now, $A-A^{\prime}=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]+\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

Let $Q=\frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

Now, $Q^{\prime}=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=-Q$

Thus, $Q=\frac{1}{2}\left(A-A^{\prime}\right)$ is a skew-symmetric matrix.

Representing A as the sum of P and Q:

$P+Q=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]+\left[\begin{array}{rrr}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]=A$

(iii)

Let $A=\left[\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]$, then $A^{\prime}=\left[\begin{array}{rrr}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]$

Now, $A+A^{\prime}=\left[\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]+\left[\begin{array}{rrr}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{rrr}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]$

Let $P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]=\left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]$

Now, $P^{\prime}=\left[\begin{array}{rrr}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]=P$

Thus, $P=\frac{1}{2}\left(A+A^{\prime}\right)$ is a symmetric matrix.

Now, $A-A^{\prime}=\left[\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]-\left[\begin{array}{rrr}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]$

Let $Q=\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]=\left[\begin{array}{ccc}0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0\end{array}\right]$

Now, $Q^{\prime}=\left[\begin{array}{ccc}0 & -\frac{5}{2} & -\frac{3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0\end{array}\right]=-Q$

Thus, $Q=\frac{1}{2}\left(A-A^{\prime}\right)$ is a skew-symmetric matrix.

Representing A as the sum of P and Q:

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

(iv)

Let $A=\left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]$, then $A^{\prime}=\left[\begin{array}{rr}1 & -1 \\ 5 & 2\end{array}\right]$

Now $A+A^{\prime}=\left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]+\left[\begin{array}{rr}1 & -1 \\ 5 & 2\end{array}\right]=\left[\begin{array}{ll}2 & 4 \\ 4 & 4\end{array}\right]$

Let $P=\frac{1}{2}\left(A+A^{\prime}\right)=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]$

Now, $P^{\prime}=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]=P$

Thus, $P=\frac{1}{2}\left(A+A^{\prime}\right)$ is a symmetric matrix.

Now, $A-A^{\prime}=\left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]-\left[\begin{array}{rr}1 & -1 \\ 5 & 2\end{array}\right]=\left[\begin{array}{rr}0 & 6 \\ -6 & 0\end{array}\right]$

Let $Q=\frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{rr}0 & 3 \\ -3 & 0\end{array}\right]$

Now, $Q^{\prime}=\left[\begin{array}{rr}0 & -3 \\ 3 & 0\end{array}\right]=-Q$

Thus, $Q=\frac{1}{2}\left(A-A^{\prime}\right)$ is a skew-symmetric matrix.

Representing A as the sum of P and Q:

$P+Q=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]+\left[\begin{array}{rr}0 & 3 \\ -3 & 0\end{array}\right]=\left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]=A$