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
$\begingroup$
$\endgroup$
0
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
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.
$\begingroup$
$\endgroup$
2
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.
-
$\begingroup$ Can you please explain your final move? how does $A$ "disappear" in your last calculation? $\endgroup$– AlanMar 27, 2016 at 19:21
-
2$\begingroup$ 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 $v_j^tv_k=\delta_{j,k}$. $\endgroup$ Mar 27, 2016 at 19:24