find-a-formula-for-the-least-squares-solution-of-a-mathbfxmathbfb-when-the-columns-of-a-are-orthonor

Name: find a formula for the least squares solution of a mathbfxmathbfb when the columns of a are orthonor
Uploaded: 2019-08-15
Duration: 1 min 50 s
Channel: Numerade
Description: Find a formula for the least-squares solution of $A \mathbf{x}=\mathbf{b}$ when the columns of $A$ are orthonormal.

Question

Find a formula for the least-squares solution of $A \mathbf{x}=\mathbf{b}$ when the columns of $A$ are orthonormal.

Angelo Rendina · Accepted Answer

So we know that if the columns of a out or focal now, then we can compute the projection onto the span of the columns of a off the given back Toby Also noting that a let&#x27;s say za and by N matrix So the production of B is given by, well, the projection onto each column of a which we denote a I we confusions given by be Doc Aye, aye. Normalized by a I don&#x27;t a I And once we know what the projection off bees, then it immediate to compute that the least square solution X hat just given by these coefficients and green. So be that a one normalized by a one, not a one all the way down to be taught a n divided by a n dot es en But now if in addition, the columns of a R or for normal, that means that hey, I got a I is just one. So all these denominators in green there just once and so we know that if the columns of a r r for normal than the least square solution except is just given by the given vector of me dot the columns of a

Angelo Rendina · Answer

we can write a genetic regression line problem in this way where we have on the first column the wise off the points. Then we have the big Matics X, which is the mother lode of coefficients, where we have in the first column all once and in the second columns the axis of the points that is my exact multiplies the Matics Beetle, which contains the parameters that we want to find a better one and bitter, too. And then we have the air or the residuals now to solve the problem. Let me just write a the metrics one over end. Well, all ones in a row. And we upset. Right now that is one over end. The first column of acts. Well, transport&#x27;s because we put it as a raw and we&#x27;re going to use this observation later. So now let&#x27;s multiply from the left by a because remember that why bar and X bar are the average is off the points and well, when we multiply from the left by A were basically obtaining the means so well different from the left. The previous equation we have a wise equal to X beat up, plus a Absalon. An hour&#x27;s observed on the left hand side. We know that it&#x27;s just why bar So the average of the wise and on the right inside will a X is going to be the vector. Well, the first entries one because the average is awful ones. And then the second column is the average of the axis. So just x bar. And then that multiplies the metrics beetle, which is better one bitter too, and then plus a Absalom. And so basically, the point is, we want to show that a absolute zero. And if we do that, the proof is over. So we know that the residuals saw Absalon is or for gonna to the columns off X or other way of writing that is that a genetic column of X. We put it, transports and we multiply by Absalon. We should get zero. But we observed that the first column Oh X is a so a upside down. He&#x27;s one over end. The first column of Ex Put as a row so transposed the multiplies Absalon and therefore that zero. Therefore from the previous page, we see that why bar is equal to be one plus beetle to x bar, which is we wanted to show

Angelo Rendina · Answer

we want to find the orthogonal projection of the victor, be onto the span of the columns of a and then Ali Square solution for the system. A x equal. Be so we observe. First of all, that the columns are they are all for. You know, you could just simply check it. And so we can use this simple method. We ride the orthogonal projection of beings or behalf he&#x27;s here and by the projection off, be onto the first column away, plus the projection off. Be onto the second column away, plus the projection off. Be onto the third column away. Now the computation of projection off be onto the columns is easy, so we get 1/3 the first column away last 14 14 over three. The second column of a minus five hours. If I thirst the third column of a putting this all together, we can just compute that their father protection behalf is given by five, two, three and six. So now, to find the least squares solution, we will need to solve this all the system a accept equal behalf, but because of the composition, we immediately see the disco efficient here. 1/3 4th in divided by three and minus 5/3 gives. Actually, the entries of exact so exact is given by 1/3 14 3rd and minus five develop battery.

Mike Gaerlan · Answer

All right. So we need to find a basis for the solution space to this differential equation. That is orthogonal. So let&#x27;s first start by finding the basis for dissolution space of we&#x27;re going to find the auxiliary equation. We shouldn&#x27;t be r squared. Plus one is equal to zero. So then now we&#x27;re going Teoh solve this So r squared is going to be the three negative one. So that means our is going to be plus or minus I. So that means that our, uh, basis for the solution space is going to be able to sign of X and co sign of X, right? So next weenie to or Thorgan allies this or check to see if it is or thought no. So we need to calculate ah b perpendicular charity or calculator North regular basis using the Gram Egemen North organization process. So when they set sign of executed view one and close by in a taxi with the V two, so remember V one perpendicular is just going to be V once is equal to sign of X, then the two per manipulator. It&#x27;s going to be two through v tu minus and then we have V one particular B two divided by the one per particular comma V one perpendicular. Be one particular. Remember, this is the projection of, uh, the two on two V one perpendicular so we&#x27;ll go ahead and calculate this here first. So V one perpendicular be to the inner product is equal to the integral from negative high to pie of sine X cause I necks like so d x So this is equal to remember ah, to sign X is equal to, uh, sign two x So this is actually going to be equal to Israel native pied a pie of 1/2 sign two x d x Now we can integrate this so integral of sine two acts is negative co sign So the negative 1/4 designed to X and this is from negative pi to pie. So hugging those numbers in we get negative 1/4 and then co sign of two pi minus ah co sign of negative two pi right co sign of two pi is going to be able to one, so that goes to one one and then this also goes to one. So we get one minus one which is equal to zero. So since this top part here is zero, um, then this whole thing just become zero. So that V two per in particular is already just equal to V two, Which is this going to be to co sign of X? Therefore, that means this is already I perpendicular set orthogonal set. So this is going to be our our agonal basis here.

Ashley Boni · Answer

So if we want to see if two vectors are or the normal, we first need to see if they&#x27;re or agonal s. So we need to take their dot product. So we have, uh, 010 and zero negative 10 So if we take the dot product where he had zero minus one close cereal, which gives us negative one, Austin says not zero. That means they are not worth organ, which means they can&#x27;t be ortho.

Jack Chen · Answer

eso in this program. We know that, um, for Patty we know that. And less recourse to t lesser because to capture him, eso if we step track this lower, Um, case in for off these three terms. So we have zero there. So he calls to t minus AM Yes. So he caused to capture and miners him? No, it divided by ah, this term for this three. Since this term is known active, so we don&#x27;t have to change the direction off BSE equality. So we have the oral Ezrin t minus and over capital and minus him. This link was toe. What? So if we just let our for you close to this number, we have t closed toe one minors off our times lower case and plus off our times kept William, which proofs the first statement in the for a second time. And if we let X equals to root off one minors off our times, you Zaban plus route off off our times to use that one. Um so x chance post times X equals two. Um, since the chance poses linear. So we have what route off one minus. Oftentimes use transposed time plus route off fraud times You one chance post times himself. And don&#x27;t forget that we have one condition Here you&#x27;re seven. So we were right out here since you seven is perpendicular to your supple one. That is you, Saban transpose times Use of one equals 20 eso This product equals two one minus often times Use your stubborn on chest post empty self plus our front times Use a bed. I use a bunch Is post times years of one This is both vectors are unit vectors So we have one minors off our process. For us, it&#x27;s one. Um, if this product is one, them is x defined in this way is also a unit vector. Now X transpose a X equals two Root off What mine is Lambda U N transpose plus route of land. You one chance force times a times myself since, um, you won again value I give actors with Begum Elio kept her in the lower case in. So we have or minus Landrieu and Chase Post, plus our route off Alfa you one chance goes times route off one mine is lambda So eight times User Bernie, close to um lower case in times the self and, uh, eight times you want enclosed the opera upper case. In times you won again. We used effect of that. You end chance post times you want because zeros away have one minus Lunda and plus off a one minus afra him plus Alfa kept him. And the by our definition before this equals to keep which proofs are last statement.

Problem

Describe all least-squares solutions of the syste…

Problem 24 Medium Difficulty

Find a formula for the least-squares solution of $A \mathbf{x}=\mathbf{b}$ when the columns of $A$ are orthonormal.

Answer

$\hat{\mathbf{x}}=A^{T} \mathbf{b}$

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Calculus 3 Educators

Lily A.

Anna Marie V.

Michael J.

Joseph L.

Calculus 3 Courses

Vectors Intro

Vector Basics - Example 1

Recommended Videos

Watch More Solved Questions in Chapter 6

Video Transcript

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$\hat{\mathbf{x}}=A^{T} \mathbf{b}$

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Lily A.

Anna Marie V.

Michael J.

Joseph L.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications