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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?

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    $\begingroup$ observe that $<Av, Aw> = <v, A^T A w> = <v, \lambda w> = 0$ $\endgroup$
    – Hayk
    May 17, 2018 at 7:31

2 Answers 2

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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.

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  • $\begingroup$ This makes sense, thank you! $\endgroup$
    – user555558
    May 17, 2018 at 7:44
  • $\begingroup$ You are welcome :) $\endgroup$
    – Bill O'Haran
    May 17, 2018 at 7:44
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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 $\lambda v^{T}w=0$.

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