# Proving Eigenvector of transpose matrix

• Nov 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.

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 $A^tA$.