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If the matrix is positive definite, then all its eigenvalues are strictly positive. Is the converse also true? That is, if the eigenvalues are strictly positive, then matrix is positive definite? Can you give example of 2x2 matrix with 2 positive eigenvalues but is not positive definite?

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In response to a (now deleted) answer: Note that "positive definite" is a term that is sometimes applied to unsymmetric matrices as well, e.g. . – J. M. Sep 9 '10 at 15:22
@J.M. can you give an example of a positive definite but asymmetric matrix? – user957 Sep 9 '10 at 16:29
$\begin{pmatrix}12&8\\\\9&10\end{pmatrix}$ – J. M. Sep 9 '10 at 16:35
For those wondering how I pulled that example out, I multiplied two SPD matrices. – J. M. Sep 9 '10 at 23:08
up vote 21 down vote accepted

I think this is false. Let $A = \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix}$ be a 2x2 matrix, in the canonical basis of $\mathbb R^2$. Then A has a double eigenvalue b=1. If $v=\begin{pmatrix}1\\1\end{pmatrix}$, then $\langle v, Av \rangle < 0$.

The point is that the matrix can have all its eigenvalues strictly positive, but it does not need to be symmetric.

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As mentioned, unsymmetric matrices can be positive definite; your example, on the other hand, shows that the answer to the titular question is no. +1. :) – J. M. Sep 9 '10 at 15:47
A 2x2 matrix $A$ with strictly positive eigenvalues but has $v^TAv <0$ is non-symmetric. – James Sep 9 '10 at 16:36
James: per Soarer's answer, yes. – J. M. Sep 9 '10 at 17:00

This question does a great job of illustrating the problem with thinking about these things in terms of coordinates. The thing that is positive-definite is not a matrix $M$ but the quadratic form $x \mapsto x^T M x$, which is a very different beast from the linear transformation $x \mapsto M x$. For one thing, the quadratic form does not depend on the antisymmetric part of $M$, so using an asymmetric matrix to define a quadratic form is redundant. And there is no reason that an asymmetric matrix and its symmetrization need to be at all related; in particular, they do not need to have the same eigenvalues.

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I agree with this. Upon reading the question, my response was, "Well, although it's possible to define positive-definiteness for an asymmetric matrix, it's not really natural to do so." Now, of course, I would like to be corrected/informed by anyone who can tell me a place where positive definite asymmetric matrices arise in a natural way... – Pete L. Clark Sep 9 '10 at 22:09
+1. However, a minor nitpick: I don't like the statement that there is no reason why a matrix and its symmetrization have to be related. There are, of course, various relationships between a matrix and its symmetrization... – alexx Sep 9 '10 at 22:10
...for example, if $A+A^T$ is negative definite, the eigenvalues of $A$ have negative real parts. – alexx Sep 9 '10 at 22:11
Pete: there is an example in the paper I linked to in the comments: discretizing certain differential operators gives rise to matrices that are the sum of a skew-symmetric matrix and the identity. The eigenvalues of this are no longer real, granted, but the matrix is PD. – J. M. Sep 9 '10 at 22:50
Pete, here is a geometric interpretation of negative definiteness which you would hopefully consider natural: a (possibly nonsymmetric) matrix $A$ is negative definite if and only if the solutions of the dynamical system $\dot{x}=Ax$ always decrease in norm, i.e. $(d/dt) ||x(t)||_2 \leq 0$. – alexx Sep 10 '10 at 3:10

The converse will be true if the matrix is diagonalizable. That's why the counter-example given by Ronaldo (edited by KennyTM) is not diagonalizable. If $B$ is diagonalizable, then $B=P^TAP$ and $y^TBy=y^TP^TAPy=x^TAx=\sum d_ix_i^2$ (where $P$ is an orthogonal matrix whose transpose is its own inverse) as in the answers by Soarer.

Spectral theorem for non-diagonalizable matrix gives rise nilpotent matrices, i.e. $$A=\sum_i (\lambda_i P_i + N_i)$$ where $P_iP_j = \delta_{ij} P_j$, $N_iP_i=N_i$ etc. ($P_i$'s are projection matrices and $N_i$'s are nilpotent matrices corresponding to eigen-values $\lambda_i$).

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Theres a problem here -- $B$ being diagonalizable does not mean that $B$ is diagonalizable by an orthogonal matrix. – John Jan 17 '15 at 6:40

As posed, the answer to the question is no, if $\mathbf A$ is not symmetric. Counterexample:

$$\mathbf A = \begin{pmatrix} 7 & 1 \\ -20 & -2\end{pmatrix}$$

with positive eigenvalues $3$ and $2$. $\mathbf A$ is not positive definite, that is, $\mathbf x^\top \mathbf A \mathbf x$ is not a positive quadratic form.

Of course, as pointed out by many, if in addition we require that $\mathbf A$ be symmetric, then all its eigenvalues are real and, moreover, $\mathbf A$ is positive definite if, and only if, all its eigenvalues are positive.

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We use $\LaTeX$ here for displaying mathematical notation; you might want to consider using it for writing out mathematical expressions. – J. M. Jul 20 '12 at 0:07

True. If you consider a diagonal matrix $A = \mathrm{diag}(d_1,\cdots,d_n)$ where each diagonal entry $d_i$ is positive, then clearly it is positive definite, since $x^TAx = \sum d_ix_i^2 > 0$ unless $x = 0$ ($x_i$ are the components of vector $x$.)

Now apply spectral theorem for symmetric matrix to reduce to the diagonal case.

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For a matrix to be positive definite:

1) it must be symmetric 2) all eigenvalues must be positive 3) it must be non singular 4) all determinants (from the top left down the diagonal to the bottom right - not jut the one determinant for the whole matrix) must be positive.

If a 2x2 positive definite matrix is plotted it should look like a bowl. If the matrix is singular then it's a trough which follows the vector which takes the matrix to zero. If the sum of the cross terms is bigger than the trace then the plot is a saddle at zero.

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