Question

The$n\times n$matrix $A=\left[a_{ij}\right]$ is called a diagonal matrix if $a_{ij}=0$ when $i\ne j$.Show that the product of $n\times n$diagonal matrices is again a diagonal matrix. Give a simple rule for determining this project.

Short answer

The product of two$n\times n$diagonal matrices is a diagonal matrix.

Step-by-step solution

Step 1

Given : A is a$n\times n$diagonal matrix if $a_{ij}=0$ when$i\ne j$

To prove : The product of two$n\times n$ diagonal matrices is a diagonal matrix.

PROOF

Let A and be B diagonal matrices, By the definition of diagonal matrices :

$a_{ij}=0$ whenever $i\ne j$

$b_{ij}=0$whenever $i\ne j$

Step 2

Let $i,j\in \{1,2,.....,n\}$ such that $i\ne j$.

$\begin{array}{r}[AB{]}_{ij}=\sum _{a=1}^p \left(\left[A{]}_{ij}\right[B{]}_{qj}\right)\\ =\left[A{]}_{i\overline{i}}\right[B{]}_{ij}+\sum _{}^p \left(\left[A{]}_{ia}\right[B{]}_{q\dot{j}}\right)\end{array}$

Since $a_{ij}=0$whenever$i\ne j$

$\begin{array}{r}=\left[A{]}_{ii}\right[B{]}_{ij}+\sum _{} \left(0\bullet [B{]}_{qi}\right)\\ =\left[A{]}_{ii}\right[B{]}_{ij}+\sum _{}^p 0\\ =\left[A{]}_{iil}\right[B{]}_{ij}+0\\ =\left[A{]}_{ii}\right[B{]}_{ij}\end{array}$

Since $b_{ij}=0$whenever $i\ne j$

$={\left[A\right]}_{ii}0$

Thus then implies that every non-diagonal element of the matrix AB is 0 and thus AB is a diagonal matrix.

The product of two $n\times n$diagonal matrices is a diagonal matrix.