# Eigenvalues for a Real Symmertic Matrix [closed]

Suppose $A$ is a $100×100$ real symmetric matrix whose diagonal entries are all positive.

How do I go on showing that at least one eigenvalue of $A$ is greater than $0$?

I have absolutely no idea how to proceed. Any hints would be appreciated.

• Hint: the trace is equal to the sum of the eigenvalues.
– amd
Commented Apr 18, 2018 at 6:57

As amd suggested, since $A$ is symmetric and real, you may diagonalize it thanks to the spectral theorem. Then, the trace of the similar diagonal matrix $D$ is the sum of all the eigenvalues of $A$ and it has the same trace as $A$.
If all the eigenvalues were strictly negative, we would have $\text{Tr}(D)=\text{Tr}(A)=\sum_{i=1}^{100}\lambda_i < 0$. Since we know that $\text{Tr}(A)\geqslant 0$, at least one eigenvalue is non-negative.
• Not really actually, there is no need for diagonalization (and thus no need for $A$ to be symmetric). But I think that involving $D$ makes it easier to visualize. Commented Apr 18, 2018 at 7:14