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Find the orthogonal vectors $v_1$ and $v_2$ such that $Av_1$ and $Av_2$ are still orthogonal, where $$A = \begin{pmatrix} 1 & -1 \\ 2 & 2 \end{pmatrix}$$

What can we comment about the lengths of $Av_1$ and $Av_2$ ?

I'm stuck thinking if these $v_1$ and $v_2$ are eigenvectors of $A$. If yes how does the second condition hold, "$Av_1$ and $Av_2$ are still orthogonal"? Is it by default? Thanks in advance!

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  • $\begingroup$ If you find orthogonal eigenvectors, then $A v_1$ and $A v_2$ must still be orthogonal. I wouldn't say "by default", but "by virtue of them being eigenvectors". $\endgroup$ – Leo Aug 12 '19 at 18:55
  • $\begingroup$ Okay thanks, is there any change in method of calculating eigenvectors if I need them to be orthogonal? $\endgroup$ – visionEnthusiast Aug 12 '19 at 18:57
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If $v_1$ and $v_2$ are eigenvectors, then $Av_1=\lambda_1v_1$ and $Av_2=\lambda_2v_2$, where $\lambda_1$ and $\lambda_2$ are the corresponding eigenvalues. So if the eigenvalues are non-zero, $Av_1=\lambda_1v_1$ and $Av_2=\lambda_2v_2$ are orthogonal if and only if the eigenvectors $v_1$ and $v_2$ are already orthogonal.

In this problem, using eigenvectors will not work since they are not orthogonal.

Instead, we can solve this problem by letting $v_1=(a, b)^T$ and $v_2=(b,-a)^T$. Then $Av_1=(a-b, 2a+2b)^T$ and $Av_2=(a+b, 2b-2a)^T$. We need $Av_1$ and $Av_2$ to be orthogonal, i.e. $(a-b)(a+b)+(2a+2b)(2b-2a)=0$. Now, solve to find a solution for $a$ and $b$.

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  • $\begingroup$ Thanks, now it makes much more sense. Great explanation! $\endgroup$ – visionEnthusiast Aug 12 '19 at 19:06

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