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My problem is the following:

Let $\textbf{H}$ be a symmetric $n\times n$ matrix. Are the following claims true? Why?

a) If the vectors $\textbf{d}_1$ and $\textbf{d}_2$ and vectors $\textbf{d}_2$ and $\textbf{d}_3$ are $\textbf{H}$-conjugate, then also the vectors $\textbf{d}_1$ and $\textbf{d}_3$ are $\textbf{H}$-conjugate.

b) If the eigenvalues $\lambda_1, \lambda_2, ..., \lambda_n$ of $\textbf{H}$ are $\lambda_i\neq\lambda_j,\;$ for $i\neq j$, then the corresponding eigenvectors $\textbf{d}_1, \textbf{d}_2, ..., \textbf{d}_n$ of $\textbf{H}$ are $\textbf{H}$-conjugate among each other.

As a reminder:

The vectors $\textbf{d}_1, ..., \textbf{d}_k$ are $\textbf{H}$-conjugate if they are linearly independent and also:

$$\textbf{d}_i^T \textbf{H}\textbf{d}_j=0,\;\;\forall\; i\neq j.$$

Any hints how to get started?

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2 Answers 2

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I'm not sure a) holds as stated because it is possible that $\mathbf{d}_3 = \mathbf{d}_1$, so they are not necessarily linearly independent. Is there another condition?

To show $\mathbf{d}_i^T \mathbf{H} \mathbf{d}_j=0$ in part b), note that $$(\lambda_i \mathbf{d}_i^T) \mathbf{d}_j = \mathbf{d}_i^T \mathbf{H} \mathbf{d}_j = \mathbf{d}_i^T (\lambda_j \mathbf{d}_j).$$

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  • $\begingroup$ Hi, thank you for your help! :) in a) no, those are the only conditions given in the problem :) $\endgroup$
    – jjepsuomi
    Dec 9, 2014 at 13:53
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Hint for a: Consider $H = I$, the $3 \times 3$ identity, and let the $d_i$ be various standard basis vectors.

Hint for b: Consider $$ d_i^T H d_j = d_i \cdot (H d_j) = (Hd_i) \cdot d_j. $$ [First explain why these equalities hold, of course!]

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  • $\begingroup$ Thank you for your help :) Appreciate it! $\endgroup$
    – jjepsuomi
    Dec 9, 2014 at 13:53

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