Product of Diagonal Matrices is Diagonal

Theorem
Let $A$ and $B$ be $n \times n$ diagonal matrices.

Then the matrix product $A B$ is an $n \times n$ diagonal matrix.

Further:


 * $\paren {A B}_{i j} = \begin {cases} \paren A_{i i} \paren B_{i i} & i = j \\ 0 & i \ne j \end {cases}$

Proof
We have:


 * $\ds \paren {A B}_{ij} = \sum_{k \mathop = 1}^n \paren A_{i k} \paren B_{k j}$

Since $A$ and $B$ are diagonal:


 * $\paren A_{i k} = 0$ for $i \ne k$,

and:


 * $\paren B_{k j} = 0$ for $k \ne j$.

If $i \ne j$, for each $k$ we either have $i \ne k$ or $k \ne j$, so:


 * $\paren A_{i k} \paren B_{k j} = 0$ for each $1 \le k \le n$.

Hence if $i \ne j$:


 * $\ds \paren {A B}_{i j} = \sum_{k \mathop = 1}^n \paren A_{i k} \paren B_{k j} = 0$

so $A B$ is diagonal.

Let $i = j$.

Then, if $k \ne i$ we have:


 * $\paren A_{i k} \paren B_{k j} = \paren A_{i k} \paren B_{k i} = 0$

so:


 * $\ds \sum_{k \mathop = 1}^n \paren A_{i k} \paren B_{k j} = \paren A_{i i} \paren B_{i i}$

We can conclude:


 * $\paren {A B}_{i j} = \begin {cases} \paren A_{i i} \paren B_{i i} & i = j \\ 0 & i \ne j \end {cases}$