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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 $2 \times 2$ matrix with $2$ positive eigenvalues but is not positive definite?

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    $\begingroup$ 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. dx.doi.org/10.1016/0024-3795(79)90122-8 . $\endgroup$ Sep 9, 2010 at 15:22
  • $\begingroup$ @J.M. can you give an example of a positive definite but asymmetric matrix? $\endgroup$
    – user957
    Sep 9, 2010 at 16:29
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    $\begingroup$ $\begin{pmatrix}12&8\\\\9&10\end{pmatrix}$ $\endgroup$ Sep 9, 2010 at 16:35
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    $\begingroup$ For those wondering how I pulled that example out, I multiplied two SPD matrices. $\endgroup$ Sep 9, 2010 at 23:08
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    $\begingroup$ @J.M. Your example can also be seen as a small perturbation of a symmetric matrix with positive eigenvalues. Small perturbation keeps the eigenvalues positive. $\endgroup$
    – K1.
    Feb 10, 2016 at 2:48

7 Answers 7

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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 follow that it is positive definite.

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    $\begingroup$ 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. :) $\endgroup$ Sep 9, 2010 at 15:47
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    $\begingroup$ A 2x2 matrix $A$ with strictly positive eigenvalues but has $v^TAv <0$ is non-symmetric. $\endgroup$
    – James
    Sep 9, 2010 at 16:36
  • $\begingroup$ James: per Soarer's answer, yes. $\endgroup$ Sep 9, 2010 at 17:00
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    $\begingroup$ Put another way, positive eigenvalues can still "flip" vectors. $\endgroup$
    – user541686
    Mar 24, 2022 at 7:49
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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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    $\begingroup$ 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... $\endgroup$ Sep 9, 2010 at 22:09
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    $\begingroup$ +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... $\endgroup$
    – morgan
    Sep 9, 2010 at 22:10
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    $\begingroup$ ...for example, if $A+A^T$ is negative definite, the eigenvalues of $A$ have negative real parts. $\endgroup$
    – morgan
    Sep 9, 2010 at 22:11
  • $\begingroup$ 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. $\endgroup$ Sep 9, 2010 at 22:50
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    $\begingroup$ 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. $(d/dt) ||x(t)||_2 \leq 0$. $\endgroup$
    – morgan
    Sep 10, 2010 at 3:10
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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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    $\begingroup$ We use $\LaTeX$ here for displaying mathematical notation; you might want to consider using it for writing out mathematical expressions. $\endgroup$ Jul 20, 2012 at 0:07
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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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    $\begingroup$ Theres a problem here -- $B$ being diagonalizable does not mean that $B$ is diagonalizable by an orthogonal matrix. $\endgroup$
    – John
    Jan 17, 2015 at 6:40
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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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In general, the converse is not true but if we assume our matrix M to be Hermitian then the converse becomes also true. By assuming our matrix to be Hermitian we have that it is unitary diagonalizable which is used to prove that it is Positive definite.

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    $\begingroup$ This is correct and, in fact, it is the best answer. Downvoting without commenting the supposed mistake should be considered bad practice. $\endgroup$
    – FCardelle
    Jun 30, 2022 at 7:40
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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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