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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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    Two special cases of what Pierre-Yves is explaining is the similarity transformation which keeps the eigenvalues as they are (for example diagonalization of a matrix) and congruence transformation which might change the location of the eigenvalues but preserve the inertia (the number of positive, zero and negative eigs) of the matrix. (i.e. Sylvester's law of inertia). Therefore when $PP^T = I = PP^{-1}$ you have the Schur decomposition given as an answer below.2011-08-22

3 Answers 3

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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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    Indeed, symmetric matrices admit a partial ordering and as you point out, we can not conclude if there are complex eigenvalues.2011-08-23
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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 $