# Prove that the multiples of two orthogonal eigenvectors with a matrix are also orthogonal

I'm having trouble with this problem, I'm lost as to how to approach it.

Let $A$ be an $m\times n$ matrix, and suppose $\mathbf{v}$ and $\mathbf{w}$ are orthogonal eigenvectors of $A^TA$. Show that $A\mathbf{v}$ and $A\mathbf{w}$ are orthogonal.

I can prove that the transpose of $A$ has the same eigenvalues as $A$, but I'm unsure how that might be helpful as they can have different eigenvectors. Any advice?

• observe that $<Av, Aw> = <v, A^T A w> = <v, \lambda w> = 0$ – Hayk May 17 '18 at 7:31

Assuming the inner product is $u^Tv$: $$(Av)^T(Aw)=v^TA^TAw=\lambda v^Tw=0$$ So $Av$ and $Aw$ are orthogonal, notice how you need not use the fact that $v$ is an eigenvector.
Inner product of $Av$ and $Aw$ is $(Av)^{T}Aw=v^{T}A^{T}Aw=v^{T}\lambda w$ where $lambda$ is the eigen value corresponding to $w$. Hence the inner product is $\lambda v^{T}w=0$.