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

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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.

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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?