Answer to Question #146298 in Matrix | Tensor Analysis for Promise Omiponle

Question

Then n x n matrix A= [aij] is called a diagonal matrix if aij= 0 when i =/= j. Verify that the product of two n x n diagonal matrices is again a diagonal matrix, and give a simple rule for determining this product.

Answer 1

let:

$A = \begin{bmatrix} a_{11} & 0 & 0 & \ldots & 0 \\ 0 & a_{22} & 0 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & \ldots & 0 & 0 & a_{nn} \end{bmatrix}$

$B = \begin{bmatrix} b_{11} & 0 & 0 & \ldots & 0 \\ 0 & b_{22} & 0 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & \ldots & 0 & 0 & b_{nn} \end{bmatrix}$

$C = \begin{bmatrix} c_{11} & c_{12} & \ldots & c_{1n} \\ c_{21} & c_{22} & \ldots & c_{2n} \\ \ldots & \ldots & \ldots & \ldots \\ c_{n1} & \ldots & c_{n2} & c_{nn} \end{bmatrix}$

Then:

$c_{11} = a_{11} \cdot b_{11} + 0 \cdot 0 + \ldots + 0 \cdot 0 = a_{11} \cdot b_{11} \\ c_{12} = a_{11} \cdot 0 + 0 \cdot b_{22} + \ldots + 0 \cdot 0 = 0 \\ \ldots \\ c_{1n} = a_{11} \cdot 0 + 0 \cdot 0 + \ldots + 0 \cdot b_{nn}= 0 \\$

$c_{21} = 0 \cdot b_{11} + a_{22} \cdot 0 + \ldots + 0 \cdot 0 = 0 \\ c_{22} = 0 \cdot 0 + a_{22} \cdot b_{22} + 0 \cdot 0 + \ldots + 0 \cdot 0 = a_{22} \cdot b_{22} \\ \ldots \\ c_{2n} = 0 \cdot 0 + a_{22} \cdot 0 + \ldots + 0 \cdot 0 + 0 \cdot b_{nn}= 0 \\$

$\ldots \\ \ldots \\ \ldots \\$

$c_{n1} = 0 \cdot b_{11} + 0 \cdot 0 + \ldots + a_{nn} \cdot 0 = 0 \\ c_{n2} = 0 \cdot 0 + 0 \cdot b_{22} + \ldots + a_{nn} \cdot 0 = 0 \\ \ldots \\ c_{nn} = 0 \cdot 0 + \ldots + 0 \cdot 0 + a_{nn} \cdot b_{nn}= a_{nn} \cdot b_{nn} \\$

It is shown that after multiplication of matrix A and B we receive matrix C which have non-zero diagonal elements, namely: $c_{ij} = 0, i \ne j, c_{ii} \ne 0,i=1,2,\ldots,n$ . So C is diagonal matrix.

And formula of multiplication of two diagonal matrices is:

$A \cdot B = C \\ c_{ii} = a_{ii} \cdot b_{ii}, i = \overline{1,n} \\ c_{ij} = 0, i \ne j$