# Prove that a symmetric matrix with a positive diagonal entry has at least one positive eigenvalue

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

## 2 Answers

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.

• what book do you recommend to learn this stuff? Commented Mar 27, 2016 at 18:44
• Serge Lang's Linear Algebra is good. Peter Petersen's book isn't bad either and it is available for free on his website. Commented Mar 27, 2016 at 18:49

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.

• Can you please explain your final move? how does $A$ "disappear" in your last calculation?
– Alan
Commented Mar 27, 2016 at 19:21
• 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 $v_j^tv_k=\delta_{j,k}$. Commented Mar 27, 2016 at 19:24