I'm trying to prove that $A^+A$ is an orthogonal projection on the column space of $A^*$ (the conjugate transpose of A), where $A^+$ is the Moore-Penrose matrix defined by $A^+=R^{-1}\overline{Q^T}$ (Q and R are obtained by applying QR decomposition to A).
However, there appears to be a problem: $A^+A=R^{-1}\overline{Q^T}QR$. Since $\overline{Q^T}Q=I$, $A^+A=R^{-1}R=I$. Can the matrix of the orthogonal projection be the identity matrix?
There are two kinds of QR decomposition in popular use. The first one is the "full" one: here $A$ is $m \times n$, $Q$ is $m \times m$, and $R$ is $m \times n$. The second is the "economy" one: $A$ is $m \times n$, $r=\operatorname{rank}(A)$, $Q$ is $m \times r$, and $R$ is $r \times n$.
In the full case, $R$ has the same size as $A$, so it can only be invertible if $A$ is square. Moreover $R$ will not be invertible unless $A$ is invertible. When $A$ is not invertible, $R$ will have some zero rows, which tell you "to put together the columns of $A$, you don't need to use these columns of $Q$". Anyway, your formula for the pseudoinverse using the full QR only makes sense if $A$ is invertible, so this is a bad way to define the pseudoinverse.
The economy case is the better way to define the pseudoinverse. Here $Q^T Q$ is indeed $I_r$, but $R$ will be invertible if and only if $r=n$. In this case $A^+ A$ is indeed $I_n$. In other words, $A^+$ is a left inverse if $A$ has linearly independent columns. (Similarly, it is a right inverse if $A$ has linearly independent rows.)
By the way, the best way to define the pseudoinverse is through the economy SVD, because then it only requires transposes (or conjugate transposes) and proper inverses.