Product of Diagonal Matrices is Diagonal
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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:
- $\paren {A B}_{i j} = \begin {cases} \paren A_{i i} \paren B_{i i} & i = j \\ 0 & i \ne j \end {cases}$
$\blacksquare$