"Let $A,M \in ,M_n(K)$ where $A$ is a symmetrical matrix $K$ is a field. If $A =^tMM$ then A is semi-definite positive."

I have seen similar questions asked but each considers either the real vector spaces or the invertible matrices.

I know if that if $A, M \in M_n(\Bbb R)$ then $^t(Mx)(Mx)\ge 0$ for every $v \in V$ but if we consdier complex numbers that statament might not hold true. Also M is not necessarly an invertible element so $A$ and $I_n$ are not necessarly congruent in this case.

Is this statament true, and if yes, how do you prove it in the complex case?

  • $\begingroup$ How do you define a positive semi-definite matrix over a general field $K$? $\endgroup$ – 57Jimmy Apr 3 '18 at 17:07
  • $\begingroup$ Oh yes, I forgot to mention I took into account only numerical fields. $\endgroup$ – reset Apr 3 '18 at 18:08

No, it's not true over the complex numbers, for the simple reason that “positive semidefinite” doesn't make sense, as $\mathbb{C}$ is not an ordered field.

Take, for instance, $n=1$ and $A=i$; then $^tAA=-1$ and, even for real $x\ne0$, $^tx\,^tAAx=-x^2<0$. For complex nonreal $x=a+bi$ we'd get $-(a+bi)^2=b^2-a^2-2abi$ and asking whether this is $\ge0$ is meaningless.

This is precisely the reason why for complex matrices one considers the Hermitian transpose (transpose and conjugate) rather than the transpose, so the product $^hx\,^hAAx$ is real (and nonnegative).

For $K=\mathbb{R}$ (or any ordered field, for that matter), the result is true, because $y_1^2+y_2^2+\dots+y_n^2\ge0$ for any $y_1,y_2,\dots,y_n\in K$.

  • $\begingroup$ So algebrically closed fields and partially ordered (also not ordered fields) are exception to this? Is there any other criteria? $\endgroup$ – reset Apr 3 '18 at 18:13
  • $\begingroup$ @reset There is no sense for non ordered fields. $\endgroup$ – egreg Apr 3 '18 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.