64

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?

Martin Sleziak
  • 53,687
user957
  • 3,387

7 Answers7

47

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.

PrincessEev
  • 43,815
Ronaldo
  • 1,263
41

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.

Qiaochu Yuan
  • 419,620
  • 2
    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 09 '10 at 22:09
  • 2
    +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... – morgan Sep 09 '10 at 22:10
  • 4
    ...for example, if $A+A^T$ is negative definite, the eigenvalues of $A$ have negative real parts. – morgan Sep 09 '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. ain't a mathematician Sep 09 '10 at 22:50
  • 2
    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. $(d/dt) ||x(t)||_2 \leq 0$. – morgan Sep 10 '10 at 03:10
  • alex, i am very interested in your example of negative definity in dynamical system. could you point to any reference which gives a further explanation? – Jack2019 May 06 '13 at 00:27
  • Interesting comment, can you clarify this part: "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."? I don't see why it only makes sense to define a quadratic form for a symmetric matrix. – Sigurd Jul 02 '18 at 18:13
  • @Sigurd: if $M$ is an antisymmetric matrix, then $x^T M x = 0$ for all $x$. – Qiaochu Yuan Jul 02 '18 at 18:36
11

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.

J. M. ain't a mathematician
  • 75,051
juan tolosa
  • 111
7

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 $\lambda_i$).

  • 2
    Theres a problem here -- $B$ being diagonalizable does not mean that $B$ is diagonalizable by an orthogonal matrix. – John Jan 17 '15 at 06:40
3

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.

J. M. ain't a mathematician
  • 75,051
1

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.

Minhaj Shakeel
  • 21
  • 1
    This is correct and, in fact, it is the best answer. Downvoting without commenting the supposed mistake should be considered bad practice. – FCardelle Jun 30 '22 at 07:40
-2

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.

shadrack
  • 1