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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?

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  • $\begingroup$ 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
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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 $\langle v_i,v_j\rangle=0$

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  • $\begingroup$ 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

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