Question

Find the symmetric and skew-symmetric parts of a matrix.

1 & 2 & 4 \\\ 6 & 8 & 1 \\\ 3 & 5 & 7 \\\ \end{matrix} \right)$$

Answer 1

Hint: We can write a matrix A as the sum of $\dfrac{A+{{A}^{T}}}{2}$ and $\dfrac{A-{{A}^{T}}}{2}$ where $\dfrac{A+{{A}^{T}}}{2}$ is symmetric part of the matrix A and $\dfrac{A-{{A}^{T}}}{2}$ is skew-symmetric part of matrix A. Now we have to find the value of ${{A}^{T}}$. ${{A}^{T}}$ represents a transpose of a matrix. A transpose of a matrix is obtained by interchanging the rows and columns of matrix A. By using ${{A}^{T}}$, we can find the symmetric and skew-symmetric parts of matrix A.

Complete step-by-step answer:
We know that a matrix can be written as the sum of symmetric matrix and skew-symmetric matrix.
We can rewrite a matrix A as follows:
$A=\dfrac{A}{2}+\dfrac{A}{2}$
Now add and subtract $\dfrac{{{A}^{T}}}{2}$on R.H.S where ${{A}^{T}}$is transpose of matrix A.
$\Rightarrow A=\dfrac{A+{{A}^{T}}}{2}+\dfrac{A-{{A}^{T}}}{2}$
Let us assume $\dfrac{A+{{A}^{T}}}{2}$ as B and $\dfrac{A-{{A}^{T}}}{2}$as C.
$\Rightarrow A=B+C....(1)$
We know that a matrix A is said to be symmetric if $A={{A}^{T}}$.
We know that
$\Rightarrow B=\dfrac{A+{{A}^{T}}}{2}$
Now we will apply transpose on both sides.

& \Rightarrow {{B}^{T}}={{\left( \dfrac{A+{{A}^{T}}}{2} \right)}^{T}} \\\ & \Rightarrow {{B}^{T}}=\dfrac{{{A}^{T}}+{{\left( {{A}^{T}} \right)}^{T}}}{2} \\\ & \Rightarrow {{B}^{T}}=\dfrac{A+{{A}^{T}}}{2} \\\ & \Rightarrow {{B}^{T}}=B \\\ \end{aligned}$$ So, we can say that B is a symmetric matrix. $$\Rightarrow B=\left( \dfrac{A-{{A}^{T}}}{2} \right)$$ Now we will apply transpose on both sides. $$\begin{aligned} & \Rightarrow {{B}^{T}}={{\left( \dfrac{A-{{A}^{T}}}{2} \right)}^{T}} \\\ & \Rightarrow {{B}^{T}}=\dfrac{{{A}^{T}}-{{\left( {{A}^{T}} \right)}^{T}}}{2} \\\ & \Rightarrow {{B}^{T}}=\dfrac{{{A}^{T}}-A}{2} \\\ & \Rightarrow {{B}^{T}}=-B \\\ \end{aligned}$$ So, we can say that B is a skew symmetric matrix. So, we can write a matrix A as the sum of a symmetric matrix and skew-symmetric matrix. From the question, we were given that $$A=\left( \begin{matrix} 1 & 2 & 4 \\\ 6 & 8 & 1 \\\ 3 & 5 & 7 \\\ \end{matrix} \right)$$ We know that transpose of a matrix is obtained by interchanging the rows and columns of matrix A. So, the transpose of A is $$\Rightarrow {{A}^{T}}=\left( \begin{matrix} 1 & 6 & 3 \\\ 2 & 8 & 5 \\\ 4 & 1 & 7 \\\ \end{matrix} \right)$$ We know that the symmetric part of a matrix is equal to $$\dfrac{A+{{A}^{T}}}{2}$$. $$\Rightarrow \dfrac{A+{{A}^{T}}}{2}=\dfrac{\left( \begin{matrix} 1 & 2 & 4 \\\ 6 & 8 & 1 \\\ 3 & 5 & 7 \\\ \end{matrix} \right)+\left( \begin{matrix} 1 & 6 & 3 \\\ 2 & 8 & 5 \\\ 4 & 1 & 7 \\\ \end{matrix} \right)}{2}$$ $$\Rightarrow \dfrac{A+{{A}^{T}}}{2}=\dfrac{\left( \begin{matrix} 2 & 8 & 7 \\\ 8 & 16 & 6 \\\ 7 & 6 & 14 \\\ \end{matrix} \right)}{2}$$ $$\Rightarrow \dfrac{A+{{A}^{T}}}{2}=\left( \begin{matrix} 1 & 6 & \dfrac{7}{2} \\\ 4 & 8 & 3 \\\ \dfrac{7}{2} & 3 & 7 \\\ \end{matrix} \right)$$ So, the symmetric part of matrix A is equal to $$\left( \begin{matrix} 1 & 6 & \dfrac{7}{2} \\\ 4 & 8 & 3 \\\ \dfrac{7}{2} & 3 & 7 \\\ \end{matrix} \right)$$. We know that the skew-symmetric part of a matrix is equal to $$\dfrac{A-{{A}^{T}}}{2}$$. $$\Rightarrow \dfrac{A-{{A}^{T}}}{2}=\dfrac{\left( \begin{matrix} 1 & 2 & 4 \\\ 6 & 8 & 1 \\\ 3 & 5 & 7 \\\ \end{matrix} \right)-\left( \begin{matrix} 1 & 6 & 3 \\\ 2 & 8 & 5 \\\ 4 & 1 & 7 \\\ \end{matrix} \right)}{2}$$ $$\Rightarrow \dfrac{A+{{A}^{T}}}{2}=\dfrac{\left( \begin{matrix} 0 & -4 & 1 \\\ 4 & 0 & -4 \\\ -1 & 4 & 0 \\\ \end{matrix} \right)}{2}$$ So, skew-symmetric part of matrix A is equal to $$\dfrac{\left( \begin{matrix} 0 & -4 & 1 \\\ 4 & 0 & -4 \\\ -1 & 4 & 0 \\\ \end{matrix} \right)}{2}$$. Note: Some students have a misconception that that a matrix A said to be symmetric if $$A=-{{A}^{T}}$$ and that a matrix A said to be skew-symmetric if $$A={{A}^{T}}$$. If this misconception is followed, then the symmetric part of A is $$\dfrac{\left( \begin{matrix} 0 & -4 & 1 \\\ 4 & 0 & -4 \\\ -1 & 4 & 0 \\\ \end{matrix} \right)}{2}$$ and skew-symmetric part of A is $$\left( \begin{matrix} 1 & 6 & \dfrac{7}{2} \\\ 4 & 8 & 3 \\\ \dfrac{7}{2} & 3 & 7 \\\ \end{matrix} \right)$$. This is a very big mistake. So, students should have a clear view on symmetric and non-symmetric matrices.