0
$\begingroup$

There is a practice problem for Chapter 3 of Prof. Strang's famous open linear algebra course (problem 7 from problem set 3). It seems the answer is incorrect, and it reveal's something I have a question about.

$$R=\left [ \begin{array}{} I & F \\ 0 & 0 \end{array} \right ]$$

$R$ is the reduced row echelon form (rref) of some matrix $A$. $F$ represents the columns of free variables and could take any values. The question is to show that the rref of $R^TR$ is $R$. Here is the solution, I think the frist part is incorrect.

enter image description here

That doesn't seem right to me, I think

$$R^TR=\left [ \begin{array}{} I & F \\ F^T & F^TF \end{array} \right ]$$

But, just working some examples, the second part I still find correct: $rref(R^TR)=R$. So finally to my question, how can I show this last identity, going from $R^TR$ to $rref(R^TR)=R$ with matrix algebra? So I see how the identity matrix clears out the lower left corner, $F^T$, but I am not seeing how this necessarily converts the bottom right, $F^TF$, to $0$.

$\endgroup$

1 Answer 1

2
$\begingroup$

Take $$C = \begin{bmatrix} I & 0\\-F^T & I \end{bmatrix}.$$ Compute $CR^TR$. This is called elementary BLOCK row operations. It can also be written as a string of elementary row operations.

$\endgroup$
4
  • $\begingroup$ I would think the purpose of this is to get $CR^T=I$ however I am not seeing that, and i am not seeing that $CR^TR=R$. $\endgroup$ Jan 10, 2020 at 3:38
  • $\begingroup$ Have you actually done the operation $C(R^TR)$? $\endgroup$ Jan 10, 2020 at 3:41
  • $\begingroup$ Sorry, my bad. I changed $C$. $\endgroup$ Jan 10, 2020 at 3:46
  • $\begingroup$ OK yes now I see it, I should have known. It is actually just 1 step of elimination just like the case of a 2x2 matrix. Thanks for the help! $\endgroup$ Jan 10, 2020 at 3:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.