$A$ has full rank iff $A^H A$ is invertible Let $A \in \mathbb{K}^{m,n}$ be a matrix. How to show that $\text{rank}(A) = n$ if and only if the matrix $A^HA$ is invertible?

Unsure of your notation/assumptions, but here's a hint: 


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*For real matrices, $$\text{rank}(A^*A)=\text{rank}(AA^*)=\text{rank}(A)=\text{rank}(A^*)$$

*For complex matrices, $$\text{rank}(A^*A)=\text{rank}(A)=\text{rank}(A^*)$$


Mouse over for more after you've pondered it a bit:

 $\text{rank}(A)=n\iff \text{rank}(A^*A)=n\iff A^*A\text{ has full rank}\iff A^*A\text{ invertible}$

