# Kronecker's $\delta$ and diagonal matrices

Let $A$ be an $n\times n$ matrix.
Prove that $A$ is a diagonal matrix if and only if $A_{ij}=\delta_{ij} A_{ij}$ for all $i$ and $j$.

$\delta_{ij}$ is kronecker delta.
I think that statement is obvious, and ironically because of that I'm not sure how to prove that!
I think it's trivial by the definition of kronecker delta, but just saying "it is trivial by the definition of kronecker delta" can be a complete proof?
Or should I say more detailed one? (I'm not sure how to explain it more specific.)

• How is "diagonal matrix" defined? – dezign Apr 1 '13 at 5:08

$A$ is diagonal iff $i\ne j \implies A_{ij}= 0$.
$i\ne j \implies \delta_{ij}=0$, and $\delta_{ii}=1$.
So, if $A$ is diagonal then for $i\ne j$: $\delta _{ij}A_{ij}=0\cdot 0=0=A_{ij}$, and for $j=i$: $\delta_ {ii}A_{ii}=1\cdot A_{ii}=A_{ii}$. So, $\delta _{ij}A_{ij}=A_{ij}$.
In the other direction, if $\delta _{ij}A_{ij}=A_{ij}$, then for all $i\ne j$: $A_{ij}=\delta_{ij}A_{ij}=0\cdot A_{ij}=0$, so $A$ is diagonal.