# Singular values, orthogonal vectors, eigenvectors

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!

• 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". – Leo Aug 12 '19 at 18:55
• Okay thanks, is there any change in method of calculating eigenvectors if I need them to be orthogonal? – visionEnthusiast Aug 12 '19 at 18:57

## 1 Answer

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 $$b$$.

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