$A$ 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?
 A: As pointed out in the comments by Franco, you need $m \ge n$.
Under this assumption, for real and complex matrices, you could argue based on the (truncated) singular value decomposition.
Let $\mathbf{A} = \mathbf{U}\mathbf{\Sigma}\mathbf{V}^{H}$ be the singular value decomposition of $\mathbf{A}$: $\mathbf{U}$ and $\mathbf{V}$ are matrices with orthonormal columns and $\mathbf{\Sigma}$ is an $n \times n$ diagonal matrix with real entries. 
Recall that the number of nonzero entries in $\mathbf{\Sigma}$ is equal to the rank of $\mathbf{A}$.
Then,
$$
\mathbf{A^{H}}\mathbf{A}
=
\mathbf{V}\mathbf{\Sigma}^{T}\mathbf{U}^{H}
\mathbf{U}\mathbf{\Sigma}\mathbf{V}^{H}
=
\mathbf{V}\mathbf{\Sigma}\mathbf{\Sigma}\mathbf{V}^{H}
=
\mathbf{V}\mathbf{\Sigma}^{2}\mathbf{V}^{H}.
$$
Note that $\mathbf{A^{H}}\mathbf{A}$ is a symmetric matrix, and that $\mathbf{V}\mathbf{\Sigma}^{2}\mathbf{V}^{H}$ is its eigenvalue decomposition:
the real eigenvalues of $\mathbf{A}^{H}\mathbf{A}$ are on the diagonal of $\mathbf{\Lambda} = \mathbf{\Sigma}^{2}$.
Finally, recall that the rank of $\mathbf{A}^{H}\mathbf{A}$ is equal to the number of its nonzero eigenvalues. 
Since the number of nonzero entries in $\mathbf{\Lambda}$ is exactly equal to the number of nonzero entries in $\Sigma$, we conclude that
$$
\text{rank}(\mathbf{A^{H}}\mathbf{A})
=
\text{rank}(\mathbf{A}).
$$
Finally, note that an $n \times n$ matrix (like $\mathbf{A^{H}}\mathbf{A}$) is invertible if and only if its rank is equal to $n$.
A: Unsure of your notation/assumptions, but here's a hint: 


*

*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:

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

