Problem set 3 from MIT 18.06 in Spring 2010 (solutions on OCW) includes the following exercise (which is problem 27 in Section 3.3 of Gilbert Strang, Introduction to Linear Algebra, 4th ed. 2009):
Suppose $R$ is $m$ by $n$ of rank $r$, with pivot columns first: \begin{align} \begin{bmatrix} I & F \\ 0 & 0 \end{bmatrix} . \end{align} [Moderator remark: Here, $R$ is supposed to be a matrix over the real numbers.]
(a) What are the shapes of those four blocks?
(b) Find a right-inverse $B$ with $RB = I$ if $r = m$.
(c) Find a left-inverse $C$ with $CR = I$ if $r = n$.
(d) What is the reduced row echelon form of $R^T$ (with shapes)?
(e) What is the reduced row echelon form of $R^T R$ (with shapes)?
Prove that $R^T R$ has the same nullspace as $R$. Later we show that $A^T A$ always has the same nullspace as $A$ (a valuable fact).
I do not understand the solution for question e, in particular the matrix block multiplication. I do not understand where the 0 matrix block comes from. When I perform the multiplication, I come up with (F transpose) time (F) instead...
Let me rename the $r \times r$ identity matrix $I$ as $I_r$. Then, \begin{align} R = \begin{pmatrix} I_r & F \\ 0 & 0 \end{pmatrix} \end{align} (in block form), thus \begin{align} R^T = \begin{pmatrix} I_r^T & 0^T \\ F^T & 0^T \end{pmatrix} = \begin{pmatrix} I_r & 0 \\ F^T & 0 \end{pmatrix} \end{align} (in block form). Hence, the rule for multiplying block matrices yields \begin{align} R^T R = \begin{pmatrix} I_r I_r + 0 \cdot 0 & I_r F + 0 \cdot 0 \\ F^T I_r + 0 \cdot 0 & F^T F + 0 \cdot 0 \end{pmatrix} = \begin{pmatrix} I_r & F \\ F^T & F^T F \end{pmatrix} \end{align} ... and you're right, this is not what the solutions claim. I think there is a typo in the solutions, because the matrix $\begin{pmatrix} I_r & F \\ F^T & 0 \end{pmatrix}$ usually has rank larger than $r$, while $R^T R$ has rank $r$. In such cases, I would normally recommend messaging the author of the solutions to point out the mistake, but I have no idea how easy it is to update an old OCW course (my impression is that it involves some indirections).
Now, how do we get the rref of $R^T R$ from this? Well, we can get it in a simpler way: It is easy to see that $R^T = \begin{pmatrix} I_r & 0 \\ F^T & 0 \end{pmatrix}$ is row-equivalent to $X := \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}$, because $R^T = \underbrace{\begin{pmatrix} I_r & 0 \\ F^T & I_r \end{pmatrix}}_{\text{an invertible $m \times m$-matrix}} X$. Hence, $R^T R$ is row-equivalent to $X R = \begin{pmatrix} I_r & F \\ 0 & 0 \end{pmatrix}$, which is already a reduced row echelon matrix. This is precisely the rref claimed by Strang.