Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 11 down vote accepted

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.

share|cite|improve this answer
    
Can you please explain your final move? how does $A$ "disappear" in your last calculation? – Alan Mar 27 at 19:21
2  
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 $v_j^tv_k=\delta_{j,k}$. – user296113 Mar 27 at 19:24

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.

share|cite|improve this answer
2  
what book do you recommend to learn this stuff? – Carry on Smiling Mar 27 at 18:44
2  
Serge Lang's Linear Algebra is good. Peter Petersen's book isn't bad either and it is available for free on his website. – User8128 Mar 27 at 18:49
    
Thanks ${}{}{}{}{}$ – Carry on Smiling Mar 27 at 18:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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