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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.

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    $\begingroup$ Hint: the trace is equal to the sum of the eigenvalues. $\endgroup$
    – amd
    Apr 18, 2018 at 6:57

2 Answers 2

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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.

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  • $\begingroup$ +1 Nicely done. $\endgroup$ Apr 18, 2018 at 7:12
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    $\begingroup$ 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. $\endgroup$ Apr 18, 2018 at 7:14
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You could use Gershgorin circles, if it is enough. You mark your diagonal entries on the complex plane and draw circles around them, whose radius is the sum of the absolute values of the other matrix entries in this row. The eigenvalues move now from the diaginal entries to any point in the union of these circles.

If any union lies strictly in the right half plane, you have a eigenvalue with positive real part.

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