Let $A$ be a symmetric $n \times n$ real matrix, $n \geq 4,$ and let $v_1, \dots, v_4 \in \mathbb{R^n}$ be non zero vectors. Suppose $Av_i = (2i-1)v_i$ for all $1 \leq i \leq 4.$ Prove that $v_1 + 2v_2$ is orthogonal to $3v_3 + 4v_4.$

Clearly $v_i$ are eigenvectors of A, each with distinct eigenvalues. Assuming one can show that the eigenvectors of a symmtric matrix can be chosen orthogonal, then is the proof trivial from there? Just compute the dot product, $(v_1 + 2v_2, 3v_3 + 4v_4) = 3(v_1,v_3) + 4(v_1,v_4) + 6(v_2,v_3) + 8(v_2,v_4) = 0,$ by orthogonality.

Is this the smartest way/fully correct way to do this?

  • $\begingroup$ I don't know about the smartest, but I think that it's certainly the way that most people would do it. In case this was an implicit part of your question: to show orthogonality of eigenvectors of a symmetric matrix associated to distinct eigenvalues, just compute the inner product $(A v, w)$ in two different ways. $\endgroup$ – LSpice Jul 12 '16 at 18:27

One has

$$\lambda_i\cdot \langle v_i,v_j\rangle=\langle Av_i,v_j\rangle=\langle v_i,Av_j\rangle=\lambda_j\cdot\langle v_i,v_j\rangle$$

This means

$$\left(\lambda_i-\lambda_j\right)\cdot\langle v_i,v_j\rangle=0$$

Which leads for $i\neq j$ and distinct eigenvalues to the required orthogonality $\langle v_i,v_j\rangle=0$

  • $\begingroup$ Yes this part is all good and well, thank you... I guess I was asking if my reasoning was valid. I was surprised by the simplicity of the question being on a practice test filled with much more tricky questions. $\endgroup$ – Merkh Jul 12 '16 at 19:27
  • $\begingroup$ Yes of course it is! Just wanted to write down the proof for completeness $\endgroup$ – marwalix Jul 12 '16 at 21:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.