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The solution that is not adding up is in the link below. Specifically, the answer provided by Chad. I really enjoyed his answer but couldn't comment (not enough reputation) on one aspect that was unclear.

Difference between orthogonal projection and least squares solution

So here is my understanding of his answer:

  1. In Chad's scenario, we had a no solution or an inconsistent system as b is not in the Column space of A. Therefore, we are solving for Ax = p instead.
  2. You use the projection formula to find p.
  3. The p vector is in the column space of A ?
  4. I can now solve the system via R-REF and get a solution set. This is where the problem is. For X1, I get a Column Vector of 1/2 and 0 not 1 and 0. Is there a possibility he made a mistake? I double checked using this site: https://matrix.reshish.com/gaussSolution.php

Greetings

  • Yes you are right, the solution set should have been $\left{t \begin{bmatrix}-2 \ 1 \end{bmatrix} + \begin{bmatrix}1/2\0\end{bmatrix} : t \in \mathbb{R}\right}$. – angryavian Jul 03 '21 at 01:09
  • Thanks. One other question. When I take the dot product of p and x, it does not quite result in 0 (orthogonal). I get -1 ? I guess it does make sense as we have to approximate? –  Jul 03 '21 at 13:35
  • In general it does not even make sense to take the dot product of $p$ and $x$. When $A$ is not square, $p$ and $x$ aren't even in the same space. – angryavian Jul 03 '21 at 16:27
  • Yeah. That didn't make much sense. I ended up figuring out the vector component of b orthogonal to C(A) by subtracting b from the projection..and then when you take the dot product of the vector component of b orthogonal to C(A) with what spans the C(A)..you do get 0. –  Jul 03 '21 at 16:34
  • Yes that's right: $b-p$ is orthogonal to $C(A)$ in general. – angryavian Jul 03 '21 at 16:44

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