KonradZuse
  • KonradZuse
6.4 #2b In Exercise 2 find the least squares solution of the linear equation Ax = b. b. > A = (Matrix(3, 2, {(1, 1) = 2, (1, 2) = -2, (2, 1) = 1, (2, 2) = 1, (3, 1) = 3, (3, 2) = 1})); b = (Vector(3, {(1) = 2, (2) = -1, (3) = 1}));
Linear Algebra
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SOLVED
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jamiebookeater
  • jamiebookeater
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KonradZuse
  • KonradZuse
\[\left[\begin{matrix}2 & -2 \\ 1 & 1 \\ 3 & 1\end{matrix}\right]\] \[\left(\begin{matrix}2 \\ -1 \\ 1\end{matrix}\right)\]
KonradZuse
  • KonradZuse
now my book says to use the formula A^T A = a^T b
KonradZuse
  • KonradZuse
and then solve.... I'll post the pratice problem.

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KonradZuse
  • KonradZuse
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tkhunny
  • tkhunny
You need a most marvelous construct: \(\left(A^{t}\cdot A\right)^{-1}\cdot A^{t}\) It is a beautifu thing.
KonradZuse
  • KonradZuse
how do we get that solution of x1 and x2? Maybe I'm over tihnking it...
KonradZuse
  • KonradZuse
Hmm... That's diff from the formula they say...
tkhunny
  • tkhunny
No, it is the same. It's just the "solved" version given the conclusion that \(\left(A^{t}\cdot A\right)\) is NonSingular.
KonradZuse
  • KonradZuse
IUc... so how do I figure out x1 and x2 at the end? I'm just confused on that part... I understand how to do it if it was just 1 matrix, but a matrix * x = another matrix I'm not so sure about?
KonradZuse
  • KonradZuse
like how did they get the answers to the question I posted above?
tkhunny
  • tkhunny
Construct that delightful matrix. I get [3/7,-2/3]^T Find \(A^{t}\cdot A\), first.
KonradZuse
  • KonradZuse
The one I posted, or their example?
tkhunny
  • tkhunny
I didn't look at the piosted example. Let's go with your original problem statement.
KonradZuse
  • KonradZuse
kk
KonradZuse
  • KonradZuse
so we have \[\left[\begin{matrix}2 & -2 \\ 1 & 1 \\ 3 & 1\end{matrix}\right]\]
tkhunny
  • tkhunny
That's "A".
KonradZuse
  • KonradZuse
yup now to find A^t
tkhunny
  • tkhunny
Which we hope is trivial, except for the type-setting. :-)
KonradZuse
  • KonradZuse
A^T should be \[\left[\begin{matrix}2 & 1 & 3 \\ -2 & 1 & 1 \end{matrix}\right]\]
tkhunny
  • tkhunny
Multiply away and you will be pleasantly surprised at the result.
KonradZuse
  • KonradZuse
\[\left[\begin{matrix}14 & 0 \\ 0 & 6\end{matrix}\right]\] is what I got.
tkhunny
  • tkhunny
Side note: Be a little more careful with your notation. So far, you have used "A" and "a' interchangeably. Also, both "T" and "t" for transpose. More consistent behavior will eliminate errors.
tkhunny
  • tkhunny
Perfect. Now the inverse of that.
KonradZuse
  • KonradZuse
The book doesn't do that but okay.
tkhunny
  • tkhunny
Positive and Diagonal - Definitely NonSingular!
KonradZuse
  • KonradZuse
\[\left[\begin{matrix}6 & 0 \\ 0 & 14 \end{matrix}\right]\]
tkhunny
  • tkhunny
Fair enough. I took a look at the example. They are setting up a small system of equestions and asking you to solve the system using any method at your disposal. On method of solving such a system is hte Matrix Inverse. That is what we are doing.
tkhunny
  • tkhunny
Close! Divide by 84.
KonradZuse
  • KonradZuse
I forgot the 1/(ad- bc)
tkhunny
  • tkhunny
Remember that \(A\cdot A^{-1} = I\)
tkhunny
  • tkhunny
Yes, that's all you left off. Good work.
KonradZuse
  • KonradZuse
1/(16 * 4) - (0 * 0)
KonradZuse
  • KonradZuse
= 1/64
tkhunny
  • tkhunny
84
KonradZuse
  • KonradZuse
woop sI did 4 not 6 :p
KonradZuse
  • KonradZuse
and 16 not 14 lol...
KonradZuse
  • KonradZuse
yeah 1/84
KonradZuse
  • KonradZuse
so we don't have to take the inverse, do we?
tkhunny
  • tkhunny
ahahah Yes, \(16*4 \ne 14*6\)
tkhunny
  • tkhunny
You did already find the inverse. It is 1/84 times the 6,14 diagonal matrix.
KonradZuse
  • KonradZuse
yeah but since the book doesn't, do we have to?
tkhunny
  • tkhunny
Yes, the book does. When it introduces x1 and x2, that is exactly what it is doing. The book just isn't using matrix methods to solve the system. This strikes me as rather odd, since this is inherently a matrix problem.
KonradZuse
  • KonradZuse
hmmm... Konfused... but if you say so ;P.
tkhunny
  • tkhunny
You started with Ax = b We added the transpose: \(A^{t}\cdot A\cdot x = A^{t}\cdot b\)
KonradZuse
  • KonradZuse
mhm.
tkhunny
  • tkhunny
Now we have a system to solve. One way to solve this is to find hte inverse of \(A^{t}\cdot A\), if it has an inverse. It does.
KonradZuse
  • KonradZuse
\[\left[\begin{matrix}\frac{1}{14} & 0 \\ 0 & \frac{1}{6}\end{matrix}\right]\]
tkhunny
  • tkhunny
This gives: \((A^{t}\cdot A)^{-1}\cdot (A^{t}\cdot A)\cdot x = (A^{t}\cdot A)^{-1}\cdot A^{t}\cdot b\), using the inverse to solve the system.
KonradZuse
  • KonradZuse
seems like we are just adding it in :P
KonradZuse
  • KonradZuse
shall I solve for A^T B?
KonradZuse
  • KonradZuse
and solving the first portion yeilds the identity matrix.
tkhunny
  • tkhunny
The inverse SOLVES the system, just like substitution or elimination or whatever other method you might use to solve the system.
tkhunny
  • tkhunny
Yes, that is what hte inverse does. The left-hand-side is just 'x' when we are done. This symbolizes the solved system.
KonradZuse
  • KonradZuse
I think I'm understanding where this is going... Instead of having to row reduce....
KonradZuse
  • KonradZuse
lke a billion times easier :P
KonradZuse
  • KonradZuse
I think I've been doing it the hard way :p
KonradZuse
  • KonradZuse
\[\left(\begin{matrix}6 \\ -4\end{matrix}\right)\]
KonradZuse
  • KonradZuse
A^Tb
tkhunny
  • tkhunny
Different problems may benefit from different methods. I think you're not quite done with this one. We don't need just \(A^{t}\cdot b\), we need \(x = (A^{t}\cdot A)^{-1}\cdot A^{t}\cdot b\)
KonradZuse
  • KonradZuse
that * the inverse = \[\left(\begin{matrix}\frac{3}{7} \\ \frac{-2}{3}\end{matrix}\right)\]
tkhunny
  • tkhunny
Done! VERY good work. This was cutting-edge stuff 40 years ago. That's WAY closer to current research than your basic 300-400 year old calculus. :-)
KonradZuse
  • KonradZuse
so x1 = 3/7 and x2 = -2/3???
KonradZuse
  • KonradZuse
or am I supposed to solve for that now?!
tkhunny
  • tkhunny
That's it.
tkhunny
  • tkhunny
Just for the record, just this Friday, I did a job for my employer using exactly this technique. My "A" was (462 x 4), so I was pretty glad I didn't have to do it by hand. :-)
KonradZuse
  • KonradZuse
Damn.... Well that was easier than what the book was doing if it came out all nicely like that :) Thanks!
tkhunny
  • tkhunny
Like I said, it is a little odd that they would introduce this technique and just bail at the end, requiring you to resort to Algebra II. Maybe the author thinks you're not ready for the Inverse? Odd, indeed.
KonradZuse
  • KonradZuse
This book really SUCKS..... Sometimes it's okay....
KonradZuse
  • KonradZuse
I have another Q that no one really answered if you don't mind taking a look at it.. I just wanted to be sure I did it correctly.
tkhunny
  • tkhunny
Well, good luck with THAT (the book and author, that is)! (laughing)

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