is this always identity matrix?

Question

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.

Answer 1

According to wikipedia noting that $R^\dagger = (R^T R)^{-1} R^T$ is the pseudoinverse we see that $$RR^\dagger = I \text{ iff } R\in\mathbb R^{m\times n} \text{ and } \mathrm{rank}(R) = n$$

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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 $I$.

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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 $$ R^T R (R^T R)^{-1} R^T = R^T$$

Answer 2

1. 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.

2. 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 $n$.