Show that if A is an n x n matrix

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Homework Statement



Show that if A is an n x n matrix whose kth row is the same as the kth row of In, then 1 is an eigenvalue of A.


Homework Equations

None that I know of.



The Attempt at a Solution

I tried creating an arbitrary matrix A and set it up to to find the det(A-λI) but that got me no where.
 
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If you know that A and At have the same eigenvalues, you can work with the original definition of eigenvalue: There is a vector x such that Atx = 1x.
 


I'm still not understanding what to do with that...
 


KristenSmith said:

Homework Statement



Show that if A is an n x n matrix whose kth row is the same as the kth row of In, then 1 is an eigenvalue of A.


Homework Equations

None that I know of.



The Attempt at a Solution

I tried creating an arbitrary matrix A and set it up to to find the det(A-λI) but that got me no where.

Think about an expansion by minors along the kth row of A-λI. You want to show 1-λ is a factor of the characteristic polynomial.
 


KristenSmith said:
I'm still not understanding what to do with that...
It is easy to find a specific vector x with Atx = 1x. That shows that x is an eigenvector of At with eigenvalue 1 => At has 1 as eigenvalue => A has 1 as 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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