Chapter 2.1, Problem 5P

Explanation of Solution

a.

Symmetric matrix:

An $n\times n$ matrix $A$ is symmetric if $A=A^T$.

Now, consider a matrix $A$ of order $n\times n$ having elements $a_{ij}$, then the elements of this transpose matrix  $A^T$ will have elements $a_{ji}$.

Now the product $A{A}^T$ is defined because both the matrices are square matrices of same order and the product will also be of the same order square matrix.

If the elements of this product matrix are $b_{ij}$, then from the definition of the product of matrices we have $b_{ij}=$ Scalar product of row $i$ of $A$ and column $j$ of $A^T$. But we know that row $i$ of $A$ is same as the column $j$ of $A^T$.

To show this, consider a $2\times 2$ matrix as given below:

$A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$

Then transpose is as follows:

$A^T=\left[\begin{array}{cc}{a}_{11}& a_{21}\\ a_{12}& a_{22}\end{array}\right]$

Therefore, multiplication is,

$A{A}^T=[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}$

Explanation of Solution

b.

Symmetric matrix:

Now taking the same matrix as in part (a.), we have

$A+A^T=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]+\left[\begin{array}{cc}{a}_{11}& a_{21}\\ a_{12}& a_{22}\end{array}\right]$

  $=[\begin{array}{cc}2a_{11}& a_{12}+a_{21}\end{array}$