In an economic decomposition $A=QR$ of $A\in \mathbb{R}^{m \times n}$ - That is, $Q \in \mathbb{R} ^{m\times n}$ and $R \in \mathbb{R} ^{n\times n}$ and $m\geq n$, does $A$ being full rank imply $R$ being full rank? Or, if not, what is the connection between $rank(A)$ and $rank(R)$?
I know that $rank(A) = rank (A^T A)$, but I am not sure if it assists me in here.
On
In the product of two matrices, the ranks of the factors are upper bounds for the rank of the product. Thus if $R$ had rank smaller than $n$, then also the rank of $A$ is smaller than $n$. Conversely, if $A$ has rank $n$, the rank of $R$ must be at least $n$. And as it can not be larger, it is $n$.
This can vary, depending on the notation of your book or paper, but generally speaking, $Q$ has an orthonormal basis of $Image(A)$ as its columns and $R$ corresponds to the coefficients of the linear combinations of the vectors of this basis so one obtains the columns of $A$.
We assume that $A$ has full column rank, i.e., its columns are linearly independent (LI), therefore dim($Image(A)$) = $n$. Simply employing the Gramm-Schmidt process, we see that since the columns of $A$ are LI, one needs to carry out full $n$ steps of the Gramm-Schmidt process in order to obtain a QR decomposition of $A$. Ending the process before the $n$-th step would necessarily mean that $Image(A) \neq Image(Q)$ (e.g. because the dimensions would be different), i.e., that columns of $Q$ do not form an orthonormal basis of $Image(A)$.
This in particular means that all of the diagonal elements in $R$ are nonzero because those correspond to the normalization and we have just shown that there will be $n$ steps of the Gramm-Schmidt process - hence also the normalization is done $n$ times.
Since $R$ is $n\times n$, is upper triangular and has nonzero diagonal it has rank $n$.
This rank-revealing also works in numerical computations, see, e.g., papers on rank-revealing QR