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Thread: Linear algebra proof.

  1. #1
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    Linear algebra proof.

    Question:
    Let $\displaystyle A = [aij]$ be the n x n matrix defined by $\displaystyle a_{ii} = k$ and $\displaystyle A_{ij} = 0 $ if $\displaystyle i \neq j$. Show that if B is any n x n matrix, then $\displaystyle AB = kB.$.

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

    What step am I missing? Although it is a relatively simple proof, I am still quite new at proof-writing. Thanks in advance.
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  2. #2
    MHF Contributor

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    Re: Linear algebra proof.

    The "$\displaystyle a_{ij}B$" 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 $\displaystyle C_{ij}= \sum_{m=1}^n A_{im}B_{mj}$. Use the fact that $\displaystyle a_{ij}= 0$ if $\displaystyle i\ne j$ to reduce that sum to a single term.
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