is this always identity matrix? do you think the following matrix multiplication results in I?
$R(R^TR)^{-1}R^T$= I   or diag(I, O)
R is not necessarily  square and may not have an inverse. 
 A: According to wikipedia noting that $R^\dagger = (R^T R)^{-1} R^T$ is the pseudoinverse we see that
$$RR^\dagger = I \text{ iff } R\in\mathbb R^{m\times n} \text{ and } \mathrm{rank}(R) = n$$

In general it is false though, see this counter-example:
Let $R = \pmatrix{1 & 0}$ then
$$R (R^T R)^{-1} R^T$$
doesn't exist because the inverse of
$$R^T R = \pmatrix{1 & 0 \\ 0 & 0}$$
doesn't exist. Especially it can't be equal to $I$.

Assuming existence of $(R^T R)^{-1}$ we can write
$$I = (R^T R)^{-1} R^T R \Rightarrow R = R (R^T R)^{-1} R^T R$$
but this does NOT imply that
$$R (R^T R)^{-1} R^T = I$$
because $R$ will not be invertible in general. We do know though that
$$ R^T R (R^T R)^{-1} R^T = R^T$$
A: *

*If $R$ is square, then $R^TR$ is invertible exactly when $R$ has full rank, i.e. is invertible.  In this case the claimed identity is true.

*If $R$ is $m\times n$, then $R^TR$ and its inverse is $n\times n$, and $R(R^TR)^{-1}R^T$ is $m\times m$.  If $m>n$, then this can never equal the identity, since the rank of the expression is at most $n$.
