# Linear algebra proof.

• September 20th 2011, 07:27 PM
Pupil
Linear algebra proof.
Question:
Let $A = [aij]$ be the n x n matrix defined by $a_{ii} = k$ and $A_{ij} = 0$ if $i \neq j$. Show that if B is any n x n matrix, then $AB = kB.$.

My work:
$A = [aij]$ is not zero only when $i = j$, as $AB = kB$ would become $0(B) = kB => 0 = 0$. Therefore, $A = [a_{ii}]$ since $j = i$ and $a_{ii} = k$. Hence, $AB => a_{ii}B => kB$.

What step am I missing? Although it is a relatively simple proof, I am still quite new at proof-writing. Thanks in advance.
• September 21st 2011, 05:46 AM
HallsofIvy
Re: Linear algebra proof.
The " $a_{ij}B$" is meaningless- it is certainly not what you mean. Also that "=>" indicates that something leads from one to the other. You need to say what that "something" is, not just indicate it.

You know, of course, that if C= AB then $C_{ij}= \sum_{m=1}^n A_{im}B_{mj}$. Use the fact that $a_{ij}= 0$ if $i\ne j$ to reduce that sum to a single term.