Eigenvalue - geometric multiplicity proof

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Homework Statement



Given matrix A:

a 1 1 ... 1
1 a 1 ... 1
1 1 a ... 1
.. . .. ... 1
1 1 1 ... a

Show there is an eigenvalue of A whose geometric multiplicity is n-1. Express its value in terms of a.

Homework Equations



general eigenvalue/vector equations

The Attempt at a Solution



My problem is I'm not sure how to start it off.
I can state A is square, symmetric and hermitian so I know it has to do with one or more of those. I tried going through using a determinant but it didn't seem to work nicely, I have a feeling that it might have to do with a property of symmetric matrices but am not sure how to go about the proof (or which property)
 
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Well, first, what is the eigenvalue in question? To do that, of course, you will need to find the characteristic polynomial. I recommend starting with "1 by 1", "2 by 2", and "3 by 3" matrices to see if you can find a pattern. Do the same thing to find the eigenvectors corresponding to that eigenvalue.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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