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Let $A$ be a symmetric martix $n \times n$ such that there is some $i$ such that $a_{ii}>0$.

Prove that $A$ has a positive eigenvalue.

I have a hint which I don't how to use/check: "Check that $a_{ii}=e^t_i*A*e_i$.

Thanks,

Alan

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2 Answers 2

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If $A$ has all non-positive eigenvalues, then it is negative semidefinite so $x^t A x \le 0$ for all $x \in \mathbb R^n$. But this contradicts $e_i^t A e_i = a_{ii} > 0$. The contradiction implies that $A$ has at least one positive eigenvalue. You can check $e_i^t Ae_i = a_{ii}$ by just performing the necessary multiplication.

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    $\begingroup$ what book do you recommend to learn this stuff? $\endgroup$
    – Asinomás
    Commented Mar 27, 2016 at 18:44
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    $\begingroup$ Serge Lang's Linear Algebra is good. Peter Petersen's book isn't bad either and it is available for free on his website. $\endgroup$
    – User8128
    Commented Mar 27, 2016 at 18:49
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By contradiction assume that all the eigenvalues $\lambda_1,\ldots,\lambda_n$ of $A$ are non positive and by spectral theorem let $(v_1,\ldots,v_n)$ an orthonormal basis of eigenvectors then using the hint let $e_i=\alpha_1v_1+\cdots+\alpha_nv_n$ and then $$a_{ii}=e_i^tAe_i=\sum_{j=1}^n\lambda_j\alpha_j^2\le0$$ which is a contradiction.

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  • $\begingroup$ Can you please explain your final move? how does $A$ "disappear" in your last calculation? $\endgroup$
    – Alan
    Commented Mar 27, 2016 at 19:21
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    $\begingroup$ Since $v_j$ is an eigenvector of $A$ associated to $\lambda_j$ then $Av_j=\lambda_j v_j$ and since this basis is orthonormal then $v_j^tv_k=\delta_{j,k}$. $\endgroup$
    – user296113
    Commented Mar 27, 2016 at 19:24

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