Question

Least squares (see comments)

Answer 1

Given the system of equations Av = b, where \[A = \left[\begin{matrix}0 & 2 \\ 1 & 2\\ -1 & 1 \\ 1 &2\end{matrix}\right], b=\left[\begin{matrix}2 \\ 2 \\1\\1\end{matrix}\right]\]find the optimal solution, in the least squares sense.

Answer 2

So I sorta understand what to do, but not really >.< i might not be too much help on this one.

The idea is, the system:

\[Ax=b\]might not have a solution. infact just by eyeballing it, it looks like there isnt a solution. So what you do is project b onto the column space. Its like saying, "i know b doesnt work, so whats the closest vector that does?"

Answer 3

Now if i could only remember the projection matrix....i think its something like:
\[A^TA\]....something something....i need to grab my books, one sec lol

Answer 4

I believe this is the Projection matrix, its really long, get ready o.O:
\[A(A^TA)^{-1}A^T\]We actually need to compute that monster >.>

Answer 5

Yeah, thats what I'm looking at, at the moment, too...\[(A^TA)^{-1}A^Tb = x\]

Answer 6

is what I have..

Answer 7

i see that in my book too. doing a little reading to make sure i know what im doing lolol

Answer 8

Is \[A^T = \left[\begin{matrix}0 & 1 &-1&1 \\ 2&2&1 & 2\end{matrix}\right]\] correct??

Answer 9

yes that is correct

Answer 10

ok, what it looks like is the best fit answer, which i'll call "x hat" is:
\[\hat x=(A^TA)^{-1}A^Tb\]

Answer 11

which is what you have up there.

Answer 12

Great :)

Answer 13

I think that was all I needed - as long as I got the A^T correct, then I'm fine :) thanks, as always, joemath!

Answer 14

no prob, sry i wasnt more help >.<