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Math Help - Determinant Property

  1. #1
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    Determinant Property

    Hi, I have a pretty simple linear algebra question, I'm just a little confused as to one of the details:

    -----------------------------

    Let A be an nxn real matrix that has the property that |a(i,i)| > sum|a(i,j)| (where i is fixed). (So, in other words the magnitude of the diagonal element is greater than the sum of the magnitudes in the rest of the row.)

    Show that det(A) is non-zero.

    -----------------------------

    I started out well enough.....I assumed det(A) = 0. Thus, at least one eigenvalue of A is zero since A is similar to it's Jordan canonical form. So, we could also conclude that the rank of A < n.

    My thought was to show that one of the diagonal entries of A would have to be zero, and this would violate the property given......but that doesn't seem to work.

    I tried writing out the bottom row as a linear combination of the first n-1 rows and tried to play with some absolute values to no avail. Is that the right way to go here?
    Last edited by joeyjoejoe; June 17th 2008 at 11:56 AM.
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  2. #2
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    Quote Originally Posted by joeyjoejoe View Post
    Hi, I have a pretty simple linear algebra question, I'm just a little confused as to one of the details:

    -----------------------------

    Let A be an nxn real matrix that has the property that |a(i,i)| > sum|a(i,j)| (where i is fixed). (So, in other words the magnitude of the diagonal element is greater than the sum of the magnitudes in the rest of the row.)

    Show that det(A) is non-zero.
    the problem is simple but not very easy! it's a little bit hard to see the trick: first of all the result is true even for complex matrices.

    now suppose, on the contrary, that \det A = 0. so there exists \bold{0} \neq \bold{x}=[x_1, \ ... \ , x_n]^T \in \mathbb{C}^n such that A \bold{x}=0. let A=(a_{ij}), and

    suppose |x_{\ell}|=\max \{|x_j|: \ 1 \leq j \leq n \}. since \bold{x} \neq \bold{0}, we have |x_{\ell}| > 0. now form A \bold{x}=\bold{0}, we'll get: \sum_{j=1}^n a_{\ell j}x_j=0, which gives us:

    a_{\ell \ell}x_{\ell}=-\sum_{j \neq \ell} a_{\ell j} x_j. thus: |a_{\ell \ell}||x_{\ell}| = \left|\sum_{j \neq \ell} a_{\ell j}x_j \right| \leq \sum_{j \neq \ell} |a_{\ell j}||x_j| \leq \left(\sum_{j \neq \ell} |a_{\ell j}| \right)|x_{\ell}|. so: |a_{\ell \ell}| \leq \sum_{j \neq \ell} |a_{\ell j}|, which is a contradiction!

    therefore: \det A \neq 0. \ \ \ \square
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  3. #3
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    Thanks, it makes sense after reading it a few times. This is one of those problems where you just have to see the trick, I was getting nothing but dead ends.
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