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

Question

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 07:31 — Hayk, May 17 '18 at 07:31

score 3 · Accepted Answer · answered May 17 '18 at 07:30

3

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.

answered May 17 '18 at 07:30

Bill O'Haran

3,034

This makes sense, thank you! – May 17 '18 at 07:44
You are welcome :) – Bill O'Haran May 17 '18 at 07:44

score 1 · Answer 2 · edited May 30 '23 at 07:46

1

Inner product of $Av$ and $Aw$ is $(Av)^{T}Aw=v^{T}A^{T}Aw=v^{T}\lambda w$, where $\lambda$ is the eigenvalue corresponding to $w$. Hence, the inner product is $\lambda v^{T}w=0$.

edited May 30 '23 at 07:46

Rócherz

3,976

answered May 17 '18 at 07:32

Kavi Rama Murthy

311,013

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

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