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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.2011-08-22
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    Hi @Pierre! Why should it have at list one plus sign?2011-08-22
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    Hi! Because one of the entries in the diagonal is positive. [Do you know quadratic forms? (If I may ask.)]2011-08-22
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    It’s the kind of questions that are very easy if you know the theory, and very hard if you don’t.2011-08-22
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    I know what it is, but it's kind of inner product, right?2011-08-22
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    Exactly! The theory tells you that you always have an orthogonal basis. If you take two such basis, the diagonal entries of your two matrices will be different in general, but you'll have the same numbers of negative, zero, and positive entries. I can try to explain this more carefully, or to find an online reference, if you're interested.2011-08-22
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    Yeah, I'm familiar with that, thanks. but what does it have to do here? can you please try to explain me again what do we want to do? Thanks!2011-08-22
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    I suppose you also know that a symmetric matrix admits an orthonormal basis formed by eigenvectors.2011-08-22
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    Yeah sure I do.2011-08-22
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    If $D$ is the diagonal matrix, then $D$ represents the same endomorphism **and** the same quadratic form as $A$. Do you see why?2011-08-22
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    $D$ is the diagonal matrix of $A$ by orthonormal-diagonalazion?2011-08-22
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    Sorry, I was busy. The point is this. Let $A$ be a matrix and $P$ an invertible matrix (both $n$ by $n$). Then $B:=PAP^{-1}$ and $A$ represent the same **endomorphism**, whereas $C:=PAP^ T$ and $A$ represent the same **quadratic form**. If $P^{-1}=P^T$, then $B=C$.2011-08-22
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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

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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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    Nice answer! +12011-08-23
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    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.2011-08-23
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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:

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