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Let $A \in M_n(\mathbb R)$ be symmetric. Given that one of the entries in its diagonal is positive, prove that it has at least one positive eigenvalue.

I didn't come to any conclusion.

Thanks guys.

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Take an orthonormal basis formed by eigenvectors. The corresponding diagonal matrix will represent the same endomorphism and the same quadratic form as $A$. Since the quadratic form has at least one plus sign, we’re done. – Pierre-Yves Gaillard Aug 22 '11 at 14:13
Hi @Pierre! Why should it have at list one plus sign? – Jozef Aug 22 '11 at 14:15
Hi! Because one of the entries in the diagonal is positive. [Do you know quadratic forms? (If I may ask.)] – Pierre-Yves Gaillard Aug 22 '11 at 14:20
It’s the kind of questions that are very easy if you know the theory, and very hard if you don’t. – Pierre-Yves Gaillard Aug 22 '11 at 14:30
I know what it is, but it's kind of inner product, right? – Jozef Aug 22 '11 at 14:32
up vote 15 down vote accepted

Suppose the matrix $A$ has no positive eigenvalue, hence a negative (semi)definite matrix $A\preceq 0$ or $x^TAx \leq 0$ for all $x\in\mathbb{R}^n$.

For simplicity, assume that the positive entry is the $A_{11}$ (or upper left etc.) entry of the matrix. Then, using a vector $x = \begin{bmatrix}1 &0 &\ldots &0\end{bmatrix}^T$, we know that $x^TAx > 0$.

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Nice answer! +1 – Pierre-Yves Gaillard Aug 23 '11 at 3:47
Where have you used the hypothesis that $A$ is symmetric? I guess in going from "no positive eigenvalue" to "negative semi-definite," as there are certainly matrices that have no positive eigenvalue but are not negative semi-definite if you don't assume (something implying) all the eigenvalues are real. – Gerry Myerson Aug 23 '11 at 4:11
Indeed, symmetric matrices admit a partial ordering and as you point out, we can not conclude if there are complex eigenvalues. – user13838 Aug 23 '11 at 9:31

Let $\ell$ be the linear transformation and $q$ the quadratic form given by $A$: $$\ell(x):=Ax,\quad q(x):=x^TAx.$$ The assumption that $A$ has a positive diagonal entry means that there is a vector on which $q$ is positive.

A matrix represents $\ell$ (in an appropriate basis) iff it is of the form $PAP^{-1}$ with $P$ invertible.

A matrix represents $q$ (in an appropriate basis) iff it is of the form $PAP^T$ with $P$ invertible.

The theory tells us that, since $A$ is symmetric (*), we can pick a $P$ such that $P^{-1}=P^T$ and $PAP^T$ is a diagonal matrix $D$. In particular, the entries of $D$ are the eigenvalues of $A$. If they were all nonpositive, $q$ would be nonpositive on all vectors.

(*) See Gerry Myerson's comment to percusse's answer.

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Because $A$ is symmetric, it admits real Schur decomposition, i.e. $A= U^t.D.U$.

Now suppose $A_{1,1}$ is positive, and assume that all eigenvalues are non-positive. Then we arrive at contradiction:

$$ A_{1,1} = \sum_{k=1}^n U^t_{1,k} \lambda_k U_{k,1} = \sum_{k=1}^n \lambda_k \left( U_{k,1} \right)^2 <= 0 $$

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