Proof of Eigenvector Property with Simple Linear Algebra

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Homework Statement


Let x be an eigenvector of A with eigenvalue \lambda and suppose x is also an eigenvector of B, corresponding to the eigenvalue \lambda2. Let C = A + B. Show that x is an eigenvector of C. What is the corresponding eigenvalue?

to the eigenvalue 2


Homework Equations





The Attempt at a Solution



{\lambdaI - A} = {\lambda2I - B}

C = 2{\lambdaI - A}
C is just a linear combination of the first eigenvector so it's got the same eigenvector.

Is this enough to complete the proof?

Is the corresponding eigenvalue just twice the original eigenvalue?
 
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why not just start by examining the product:
Cx
 
Last edited:
Start like this: you know Ax = \lambda x and Bx = \lambda_{2} x from the definition of an eigenvector

Thus Cx = (A + B)x = ... and go from there. I think it will be straightforward.
 
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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