# If $AA^T$ is positive definite, is A invertible? [closed]

An answer to the question does exist at If $MM^{'}$ is positive definite, $M$ is invertible?

However, the answer uses determinants. I was wondering if there is any other way to go about it, by showing directly that the null space of $$A$$ is empty?

• Assuming that $A$ is square, then yes. Commented Jan 18 at 22:07
• Maybe use rank-nullity theorem? If the kernel is trivial, the matrix is full rank and is invertible. I am not sure if the proof of rank-nullity theorem uses determinants though. Commented Jan 18 at 22:08
• If the null space is non-trivial then $0$ is an eigenvalue so it is not positive definite. Commented Jan 18 at 22:14
• @copper.hat Isn't $A$ necessarily square for it to make sense that $AA^T$ is positive definite? Commented Jan 18 at 22:22
• @Filippo no ${}{}$. Commented Jan 18 at 22:30

Note that $$AA^T$$ positive definite means for any nonzero $$x$$,
$$\left\|A^Tx\right\|^2 = x^TAA^Tx >0,$$
meaning $$A^Tx$$ is nonzero for all $$x\neq 0$$. Therefore, $$A^T$$ is invertible, hence so is $$A = (A^{-T})^{-T}$$.
• You mean "for all non-zero $x$", right? Commented Jan 18 at 22:14
• Is there a difference between $A^T$ and $A^{-T}$? Commented Jan 18 at 22:17
• Oh it's the inverse of $A^T$, right? Commented Jan 18 at 22:20