# Proving Eigenvector of transpose matrix

• November 14th 2010, 05:42 PM
kenshin36
Proving 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.

• November 15th 2010, 12:08 AM
FernandoRevilla
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.
• November 15th 2010, 02:04 AM
Ackbeet
What ideas have you had so far?