Distinct eigenvalues

[Imum Coeli]

I was wondering if the following reasoning is correct?

If A has n distinct eigenvalues then A has n linearly independent eigenvectors.
So if A is 4 by 4 and has distinct eigenvalues, where

A*[0 1 0 1]' = [0 1 0 1]' and A*[1 0 1 0]' = [-1 0 -1 0]'

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Then [1 1 1 1]' cannot be an eigenvector because it is a linear combination of [0 1 0 1]' and [-1 0 -1 0]' ?

Thanks.

 

[eddybob123]

Yes, that is correct.

 

[Imum Coeli]

Thank you.

 

[daon2]

Think about the general problem. Suppose \(\displaystyle (v,\lambda), (w,\mu)\) are eigen-pairs for a matrix \(\displaystyle A\).

Then \(\displaystyle A(v+w) = Av+Aw=\lambda v + \mu w\). So \(\displaystyle \lambda = \mu\) if and only if \(\displaystyle A(v+w) = \lambda(v+w)\).