let A be an nxn matrix and suppose that u and v are eigenvectors of A,both with corresponding eigenvalue y.Prove that if u+v is not equal to 0n , then u+v is also an eigenvector of A with corresponding eigenvalue y

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- Dec 7th 2008, 04:53 AMmath8553eigenvector u+v
let A be an nxn matrix and suppose that u and v are eigenvectors of A,both with corresponding eigenvalue y.Prove that if u+v is not equal to 0n , then u+v is also an eigenvector of A with corresponding eigenvalue y

- Dec 7th 2008, 08:15 AMThePerfectHacker
By definition $\displaystyle A\bold{u} = y\bold{u}$ and $\displaystyle A\bold{v} = y\bold{v}$.

This means, $\displaystyle A\bold{u} + A\bold{v} = y\bold{u}+y\bold{v}\implies A(\bold{u}+\bold{v}) = y(\bold{u}+\bold{v})$

Thus, if $\displaystyle \bold{u}+\bold{v}\not = 0 $ then it is an eigenvector with eigenvalue $\displaystyle y$.