If v is an eigenvector of the matrix AA^T (A multiplied by its transpose) with eigenvalue λ, show that A^Tv is an eigenvector of A^TA.

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- Nov 14th 2010, 05:42 PMkenshin36Proving Eigenvector of transpose matrix
If v is an eigenvector of the matrix AA^T (A multiplied by its transpose) with eigenvalue λ, show that A^Tv is an eigenvector of A^TA.

Please Help >_< - Nov 15th 2010, 12:08 AMFernandoRevilla
By hypothesis there exits an scalar $\displaystyle \lambda$ such that $\displaystyle (AA^t)v=\lambda v$. Then,

$\displaystyle (A^tA)(A^tv)=A^t[(AA^t)v]=A^t(\lambda v)=\lambda(A^tv)$

That is, $\displaystyle A^tv$ is an eigenvector of $\displaystyle A^tA$.

Regards. - Nov 15th 2010, 02:04 AMAckbeet
What ideas have you had so far?