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Math Help - Prove this matrix is invertible

  1. #1
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    Prove this matrix is invertible

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

    Attempt: if A^2=0 then A is a singular matrix. A singular matrix times \alpha is still a singular matrix. so the idea is to prove that 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 = \left(\begin{array}{cc}1&1\\2&2\end{array}\right) D = \left(\begin{array}{cc}-1&0\\0&0\end{array}\right) C and D are both singular matrices.

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

    however,

    2) D + \left(\begin{array}{cc}1&0\\0&1\end{array}\right) = \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 A^2=0

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

    Try computing

    (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

    (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 A^2=0 to try and square I+\alpha A to see what that'd get me. In the result of (I+\alpha A)(I+\alpha A), the term \alpha^2A^2 of course vanishes, but the cross-term 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 \alpha A in one of the factors. That's it.
    Last edited by HappyJoe; October 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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