We know that a symmetric matrix $A$ is positive semidefinite i.e. $x^TAx \geq 0$ if and only if all its eigenvalues are nonnegative.

Now suppose I have a matrix (not necessarily symmetric) $A$, whereby all its eigenvalues are nonnegative (obviously real), is it positive semidefinite?

My hunch is yes. Because such matrix $A$ would satisfy $\lambda_{\min}(A)\|x\|^2 \leq x^TAx$. Therefore it has to be positive semidefinite.

But I have searched up and down through every linear algebra book that I have came across, virtually all of them states definition with respect to symmetric positive semidefinite matrix only.

  • $\begingroup$ As far as I know, positive semidefinite applies only to symmetric matrices, and if this is accurate then your question makes no sense $\endgroup$ – DonAntonio Jun 13 '17 at 22:08
  • $\begingroup$ @DonAntonio There are examples of asymmetric matrices such that $x^TAx \geq 0$. In some of these cases, $A$ does not even have any real eigenvalues. Then if $A$ is positive semidefinite means $x^TAx \geq 0$, then positive semidefinite can be extended to asymmetric matrices. $\endgroup$ – Olórin Jun 13 '17 at 22:09
  • $\begingroup$ The inequality $\lambda_{min}(A)\|x\|^2\leq x^TAx$ might be problematic. $\endgroup$ – Jack Jun 13 '17 at 22:10
  • $\begingroup$ In $\lambda_{\min}(A)\|x\|^2 \leq x^TAx$ the valued $\lambda_{\min}$ is the minimum eigenvalue of the symmetric part of $A$, not of the full $A$. $\endgroup$ – enzotib Jun 13 '17 at 22:30
  • $\begingroup$ @JohnathanMeek Never heard of. I've only read of that applied to symmetric or, in the complex case, Hermitian matrices. $\endgroup$ – DonAntonio Jun 13 '17 at 22:54

How about $$A=\pmatrix{1&-10\\0&2}?$$

  • $\begingroup$ I'm confused between your example and the answer here math.stackexchange.com/questions/518890/… $\endgroup$ – Olórin Jun 13 '17 at 21:57
  • $\begingroup$ @JohnathanMeek in the other answer, symmetry of the matrix seems to be implicitly assumed. Many people define positive definiteness to imply/include symmetry. $\endgroup$ – daw Jun 15 '17 at 13:56

The answer to your question is NO. (And it is worse in the complex case: if a matrix $A$ is not Hermitian, then it is impossible that $x^TAx\geq 0$ for all $x$.)

In the setting of complex vector spaces, if $x^TAx\geq 0$ for all complex vectors $x$ (your definition of semi-definite), which in particular implies that $x^TAx\in{\bf R}$, the following theorem in Axler's Linear Algebra Done Right shows that $A$ must be Hermitian (if $A$ is real, then $A$ must be symmetric):

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In the setting of real vector space, one has the following simple counterexample.

For any $(x,y)\in{\bf R}^2$, $$ (x,y)\begin{pmatrix} 0&1\\ 0&0 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}=xy. $$


Given that, for a real matrix $A$ we have $$ x^T A x=x^T A^{(S)} x $$ where $A^{(S)}$ is the symmetric part of $A$, then the character of $A$ with respect to positive definiteness, or semi-definiteness, is related only to the eigenvalues of its symmetric part.

As the examples in other answers show, a matrix could have positive eigenvalues, but its symmetric part could have a negative eigenvalue, so eigenvalues of a matrix could not be related to positive (semi)definiteness.


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