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Thread: Prove this matrix is invertible

  1. #1
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    Prove this matrix is invertible

    $\displaystyle A$ is a square matrix such that $\displaystyle A^2$=0. Prove that for any real number $\displaystyle \alpha$ the matrix $\displaystyle I+\alpha A$ is invertible

    Attempt: if $\displaystyle A^2$=0 then $\displaystyle A$ is a singular matrix. A singular matrix times $\displaystyle \alpha$ is still a singular matrix. so the idea is to prove that $\displaystyle I+A$ is an invertible matrix when A is singular.

    I'm not sure where to go from here but this is something interesting I found:

    C = $\displaystyle \left(\begin{array}{cc}1&1\\2&2\end{array}\right)$ D = $\displaystyle \left(\begin{array}{cc}-1&0\\0&0\end{array}\right)$ C and D are both singular matrices.

    1) C + $\displaystyle \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$ = $\displaystyle \left(\begin{array}{cc}2&1\\2&3\end{array}\right)$ = invertible matrix

    however,

    2) D + $\displaystyle \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$ = $\displaystyle \left(\begin{array}{cc}0&0\\0&1\end{array}\right)$ = singluar matrix

    So I believe the idea is to figure out why D is not part of the matrix we are looking for, even though it is singular, and it probably has to do with the fact that $\displaystyle A^2=0$

    Thanks!
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  2. #2
    Member HappyJoe's Avatar
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    Here's an idea:

    Try computing

    $\displaystyle (I+\alpha A)(I-\alpha A).$
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  3. #3
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    Quote Originally Posted by HappyJoe View Post
    Here's an idea:

    Try computing

    $\displaystyle (I+\alpha A)(I-\alpha A).$
    Genius (:

    Is this a common type of question. How'd you think to do that?
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  4. #4
    Member HappyJoe's Avatar
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    I'm not sure if it's common. I haven't seen it before, but it's a fun result.

    As for my approach, I was first inspired by the condition $\displaystyle A^2=0$ to try and square $\displaystyle I+\alpha A$ to see what that'd get me. In the result of $\displaystyle (I+\alpha A)(I+\alpha A)$, the term $\displaystyle \alpha^2A^2$ of course vanishes, but the cross-term $\displaystyle 2\alpha A$ doesn't, which is pretty useless. Then I just remembered that to get the cross-term out of the way, you can change the sign on $\displaystyle \alpha A$ in one of the factors. That's it.
    Last edited by HappyJoe; Oct 17th 2010 at 06:11 AM. Reason: Bad LaTeX.
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  5. #5
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    Thanks Joe! Stay happy (:
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