10
$\begingroup$

I mean in $\mathbb{C}$ positive definite matrices seem to be self-adjoint. For matrices over real vector spaces this seems to be wrong, but is it still true that they are diagonalizable?

Then everyone knows a result similar to this: When a matrix has only positive eigenvalues then it is positive definite. Is this result always true, and do we also have the converse in general?

$\endgroup$
5
  • 4
    $\begingroup$ Many authors mean positive definite and symmetric (or self-adjoint) when they write simply positive definite. It is symmetry which implies diagonalizable, so really this is a question about what you mean by positive definite. $\endgroup$
    – hardmath
    Aug 21, 2013 at 20:58
  • $\begingroup$ ah, that explains a lot, since I was wondering about some passages in a book that I found on it, great remark, thanks, but NO I do not want the matrix to be necessarily symmetric $\endgroup$
    – user66906
    Aug 21, 2013 at 21:00
  • $\begingroup$ Also, a matrix is positive definite only if its eigenvalues are all positive. A symmetric matrix with only positive eigenvalues is positive definite. $\endgroup$
    – Jonathan Y.
    Aug 21, 2013 at 21:01
  • $\begingroup$ @JonathanY.: You could of course define "positive definite" to mean that, but it is not a widely used definition. See this Question and Answers. $\endgroup$
    – hardmath
    Aug 22, 2013 at 2:45
  • $\begingroup$ @hardmath, I believe I'm using the standard definition of "positive-definite" (correct me if I'm wrong). However, if a matrix has a non-positive eigenvalue, showing that it's not positive-definite is straightforward. A real symmetric matrix, on the other hand, is normal, hence it has an orthogonal basis of eigenvectors; the road is paved to showing the other direction as well for this type of matrix. $\endgroup$
    – Jonathan Y.
    Aug 22, 2013 at 11:32

1 Answer 1

Reset to default
12
$\begingroup$

An extension of the definition of "positive definite" to non-symmetric (real) matrices is to require $v^T M v \gt 0$ for all suitable nonzero vectors $v$. However see the discussion on this older Question about the lack of a conventionally adopted definition.

For real matrices this allows for non-self-adjoint (non-symmetric) examples. However the analogous "extension" for complex matrices, $v^* M v \gt 0$ for all nonzero vectors, turns out to imply $M$ self-adjoint (Hermitian), so it gives "nothing new".

As far as diagonalizability goes, neither requiring all eigenvalues positive nor requiring positive definiteness in the sense given above is sufficient to imply similarity to a diagonal matrix. Of course similar to a diagonal matrix would be implied by a full set of distinct eigenvalues (as each eigenvalue has at least one eigenvector), and it is equivalent to a basis of eigenvectors (since the matrix representation wrt to such basis is diagonal).

Here's an example of a nonsymmetric real matrix that (a) has only the positive eigenvalue 1 and (b) satisfies the "coercivity" condition $v^T M v \gt 0$ for all nonzero (real) vectors but (c) cannot be diagonalized:

$$ M = \begin{pmatrix} 1 & 0.1 \\ 0 & 1 \end{pmatrix} $$

Note that if $v^T = (x \; y)$, then $v^T M v = x^2 + 0.1xy + y^2$, which is easily shown to be a positive definite quadratic form, and thus the second (stronger) notion of a positive definite (but nonsymmetric) real matrix holds.

$\endgroup$
1
  • 1
    $\begingroup$ Nice answer, just adding here thatt $M$ above does have a square root that's positive definite (but of course not symmetric), namely: $ sqrt(M) = \begin{pmatrix} 1 & 0.05 \\ 0 & 1 \end{pmatrix} $ $\endgroup$
    – Learning Math
    Nov 19, 2019 at 14:51

You must log in to answer this question.