# The product of a matrix with it's transpose is always semi-definite positive

"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?

• How do you define a positive semi-definite matrix over a general field $K$? – 57Jimmy Apr 3 '18 at 17:07
• Oh yes, I forgot to mention I took into account only numerical fields. – 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 $y_1,y_2,\dots,y_n\in K$.