Question

Find the symmetric and skew-symmetric parts of a matrix.
\[A=\left( \begin{matrix}
   1 & 2 & 4 \\
   6 & 8 & 1 \\
   3 & 5 & 7 \\
\end{matrix} \right)\]

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.
\[\begin{align}
  & \Rightarrow {{B}^{T}}={{\left( \dfrac{A+{{A}^{T}}}{2} \right)}^{T}} \\
 & \Rightarrow {{B}^{T}}=\dfrac{{{A}^{T}}+{{\left( {{A}^{T}} \right)}^{T}}}{2} \\
 & \Rightarrow {{B}^{T}}=\dfrac{A+{{A}^{T}}}{2} \\
 & \Rightarrow {{B}^{T}}=B \\
\end{align}\]
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.
\[\begin{align}
 & \Rightarrow {{B}^{T}}={{\left( \dfrac{A-{{A}^{T}}}{2} \right)}^{T}} \\
 & \Rightarrow {{B}^{T}}=\dfrac{{{A}^{T}}-{{\left( {{A}^{T}} \right)}^{T}}}{2} \\
 & \Rightarrow {{B}^{T}}=\dfrac{{{A}^{T}}-A}{2} \\
 & \Rightarrow {{B}^{T}}=-B \\
\end{align}\]

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.