Proving Eigenvector of transpose matrix

**kenshin36** · November 14th 2010, 05:42 PM

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 >_<

**FernandoRevilla** · November 15th 2010, 12:08 AM

By hypothesis there exits an scalar $\lambda$ such that $(AA^t)v=\lambda v$ . Then,

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

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

Regards.

**Ackbeet** · November 15th 2010, 02:04 AM

What ideas have you had so far?

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