find-an-orthogonal-basis-for-the-column-space-of-each-matrix-in-exercises-9-12-leftbeginarrayrrr1-3-

Question

Find an orthogonal basis for the column space of each matrix in Exercises $9-12$ .
$$
\left[\begin{array}{rrr}{1} & {3} & {5} \ {-1} & {-3} & {1} \ {0} & {2} & {3} \ {1} & {5} & {2} \ {1} & {5} & {8}\end{array}ight]
$$

Sean Thrasher · Accepted Answer

to find the north orginal basis for the column space at this matrix. The column space was the subspace spanned by the column vectors of the Matrix. And so, essentially, we gotta spectra subspace That is the span of X one extra extra. And I&#x27;ve labeled them like this, right? And so I want to find a new basis for you want me to be three? That is the same. So that is the basis for the same subspace w the column space. However, I want this set of actors to be orthogonal, so v one not be too. Must be 031 dot be three. Must be zero. And so must be vey to dot B three. Okay, So clearly, we&#x27;re gonna have to apply the Gram Schmidt procedure. OK to x one. Being this vector X to being this vector next three being this vector. So what does Gram Schmidt to tell us? It&#x27;s a V one is equal to x one. Me, too is equal to x to minus a subtract x two dot v one Polar mont theme Magnitude squared of everyone. Lots of the one. And so this makes X two perpendicular to be one. And now for the three attic X three, I make it perpendicular to be one. Andi, I also make a perpendicular to V two. Okay, so this right here and is what I need to use formula that I need to use. Okay, so let&#x27;s go to the next page. So we know what? Okay. So, basically, I want to find what V one is first, then find V two and then finally find V three. Well, V one is the same as X one, as we said. So I&#x27;ve written that down V two. So let&#x27;s just focus on B two. It&#x27;s this right comes okay. We need to figure out what this V two is is a vector. What? That what? The components are, by the way, have written them like this as rose. Just a safe space. Nothing else. Okay, so, as always, you should find what these guys are First, to make things a little bit easier. So what is x two dot v one? That&#x27;s the same things. X two dot x one. Right. So it&#x27;s three plus three plus five plus five. That gives you 16. What is V one more squared? That&#x27;s one plus one plus one plus one. So that&#x27;s four. And so this is the same thing is X two minus 16 by four times V one. So that&#x27;s X two minus four times being one. So let&#x27;s figure out what V to actually is. So that&#x27;s three minus 3 to 55 minus four times one minus 1011 And so that&#x27;s the same thing is three minus four minus three plus four to five, minus 45 minus four. And so that gives me minus 11211 Let me write that down minus 1121 point. Okay, so let me update. What we&#x27;ve got so far is one minus one zero&#x27;s There, there. 11 is V one. V two is minus 11211 Now the question is, what is V three? So let&#x27;s go back to what Gram Schmidt tells us. Be three is given by this. Okay, as always, let&#x27;s figure out what these quantities are, first of all, so x three don&#x27;t be one. So that&#x27;s, uh, five, five times one minus one plus two plus eight. Okay, so five minus one plus two plus eight. So that gives me 14. Okay, x three dot ve to. That&#x27;s so be careful that you don&#x27;t use X to You need to use this guy right here. So it&#x27;s minus five plus one plus six plus one plus hates. Okay, so that&#x27;s, um Okay, so let me just check. Minus why was one plus six plus two plus states? Okay, so that&#x27;s 10 plus two. That gives me 12. The square of the magnitude of the one. Well, we&#x27;ve already worked that out. That&#x27;s just four square the magnitude of the two. So one squared, plus one squared plus two squared plus one speck plus one squared. That gives us eight. Okay, so All right. So I want to be careful not to run out of too much space. So the 36 x three minus 14. Divided by. It&#x27;s a 14 divided by four. It was the one minus 12 divided by eight times. Me too. And so that gives me X three minus seven Harms. We want minus three. House beat you. Okay, so now let me actually figure out what this vector is. So now it&#x27;s going to get a little messy, so this is going to be. So let me just write it out. So one line ist ones over one minus 11211 Yep. So this is extreme minus seven. Hearts be one minus three House. Me too. All right, this is what that is. So let&#x27;s combine this into a single vector. We get five minus seven halves, plus three halves on. Yeah, just it&#x27;s down to us arithmetic at this point, arithmetic that you should be careful not to mess up because it&#x27;s really easy to mess up arithmetic. Okay, so that&#x27;s five minus 21 plus 20 to minus five and eight, minus five. And so that gives us finally 3303 It&#x27;s a 33 033 3303 Okay. And so it&#x27;s a good idea to check that the one don&#x27;t be too equals. We wound up the three because the 2.33 cause if this is Izzy, put zero. Because if this is false, then you made a mistake somewhere. Okay, All righty. So that&#x27;s not conclusion that a North Oregon or basis for the column space is given by this. It&#x27;s driven by this collection of actors. Okay, just is a note I sometimes switch between using square braces and round braces. It really doesn&#x27;t matter. All right, so that&#x27;s it. That&#x27;s this question done. I&#x27;ll see you in the next video.

Sean Thrasher · Answer

Okay, So in this question were asked to find the North organo basis for the column space of this matrix. So what is a column space? It&#x27;s simply the subspace that has spanned by the vectors that are the columns, the Matrix. So each of the column vectors Now note that I&#x27;m not saying that this thes form a basis for the column space they might not, right. They might not be linearly independent, but it turns out that it doesn&#x27;t really matter to us. Essentially, all this question really goes down. Bulls down to is Well, let w b the subspace that is spanned by these guys. Okay, this is the notation for no the spanning sets. Essentially, we just want to find a north orginal basis for W. So we just need to apply the Gram Schmidt procedure to these vectors. So I called this guy x one x to next three. We want to find a basis for W x one x two x three Such That wasn&#x27;t orthogonal set a back. Just sorry. We want to find ah basis with what we start with X one extracts through when we want to find a basis w for W V one b two b three such that this is a north agonal set of actor. So everyone Davi to zero V one of e b three envy to Davi. Three year old zero. Okay. And they form a basis for the same subspace w. So we to apply the ground stripped procedure. And I&#x27;m just going to quote, sit here, Okay? The case of three factors. So let me go to the next page. All right. I&#x27;ve written down x one x two x three as row factors. Just a safe space. So V one is going to be X one. Okay, V two is going to be x two. And then from that to make perpendicular to the one I subtract extrude everyone over. We won on squared lots of the one. The three is x three minus x three dots V one. We want not squared. Minus from that. This. Okay, so we need to figure out what the oneness. Then we&#x27;ll figure out what be too is. Then we&#x27;ll figure out what V three is. So the one that&#x27;s very easy. It&#x27;s the same as X one. So let me just write it down So what is V two? Well, as always before, you know, before trying to go in and plug in everything like that, let&#x27;s figure out what all these guys are, right? All these dot products and you know, magnitude squares are So what is x two dot v one? So we&#x27;re working with the two first. Let&#x27;s forget about feet three for now. So x two V one is equal to Well, it&#x27;s six times. Okay, so six times minus one plus minus. Eight times three plus minus, two times one plus minus. Four times one. And that&#x27;s minus six plus minus 24 minus two minus four. And so you can check that. That&#x27;s minus 36. Okay, what about this? What is this? That&#x27;s one squared plus three squared plus one squared plus one squared, which is just 12. Okay, so let&#x27;s stop. Unless actually right, then what v two is so so v two. This was the same as X one. So v two is equal to x two minus X to doubt me. One over what we want squared to be one. So this waas. So we said that this was minus 36 divided by 12 x two plus three. Lots of the one. So that&#x27;s actually figured that out. So six minus eight minus two minus four plus three times minus 1311 So that&#x27;s 31 um, one and minus one. Okay, just check that. That&#x27;s rights. That&#x27;s correct. Okay, so let me write that down. Because 311 minus one. Okay, now we need to figure out what the three is. Hey, so let&#x27;s just do that quickly. Okay? So, two, let&#x27;s go back. So v three was Let&#x27;s remind ourselves it&#x27;s x three minus extra dot ve won over. Everyone squared. This is a three x three dot B two over more B two squared. You wanna hear it sends me too. Okay, so what is x three dot v one equal to so again? I&#x27;m just figuring out what these guys are separately first. So extra dot to be one, she let me do that over here. So extra don&#x27;t be one. So that&#x27;s equal to minus six plus nine plus six minus three. Okay, so that&#x27;s six. Okay, what&#x27;s x three dogs ve to equal to extradite me to is equal to So we need this guy. So it&#x27;s 18 plus three plus six plus three. All right? Yeah. Okay, so that&#x27;s fine. So that&#x27;s 12 plus 18. So that&#x27;s equal to 30. And then much of this, everyone squared, was 12 on model SUV. Two squared. What is that equal to, um, that&#x27;s equal to Well, you can check for yourself that it&#x27;s well, three squared is so it&#x27;s the same. Yeah. So it&#x27;s equal to 12 as well. All right, so let&#x27;s get rid of this. Now, let&#x27;s substitute in what we know so far again, I should put arrows over all these guys, but I don&#x27;t to say because it gets a little bit muddled up with, you know, with writing it all out. Um, So Okay, so that&#x27;s six over six, divided by 12 times. Be one minus 30. Divided by 12 times. Me, too. Okay, so for me, me make. So Okay, so that&#x27;s going to be two x three dots. Want minus one? Happy one, minus five. Half speed, too. X three minus 1/2 G one minus five. House. You too. That&#x27;s what the three is so that Okay, so let&#x27;s try to All right. Um okay. You copy? and paste all this, but just to remind myself he never minds. Okay, so Right. So let&#x27;s actually figure out what the three actually is. So? So I&#x27;m just using. I&#x27;m just looking at this at 636 minus three minus 1/2 times minus 131 one minus five halves times 311 minus one. And that is equal to. So let me leave quite a bit of space for this. Okay? So I&#x27;m going to computed, as always, of course. Component by component six plus 1/2 minus 15. House three minus three halves, minus five halves. Six minus 1/2 minus five halfs minus three minus 1/2 plus five lives. So that is the same thing at six. So minus seven three minus four six minus three minus three plus two. So that gives you minus one minus 13 minus one. Let me write that. That minus one minus 13 minus one. Okay, just do a quick check that these are indeed orthogonal. So figure out the one. Don&#x27;t be too. We want about 33 and beef 233 And they all are indeed equal to zero. And quickly shake. That&#x27;s so our conclusion is that we found an orthogonal basis for the column space after we&#x27;ve applied the Gram Schmidt procedure. So therefore, on orthogonal basis for the column, space is minus 1311 311 minus one minus one minus 13 minus one. Okay, so just a quick note for this to be a basis as to span and be linearly independent, right? So how could you be sure that is that these are linearly independent because we didn&#x27;t start with a set that we were guaranteed were linearly independent. We didn&#x27;t bother shaking that. Well, each of these vectors are non zero, right? And they&#x27;re all our signal to each other. It turns out that if you have a bunch of non zero vectors that are all orthogonal to each other, they indeed have to be linearly independent. This is just a side note, right? We&#x27;ve already finished the question. Um, so if you started out with a set of actors that were linearly dependent than you actually end up with, uh, some of the vectors ending up being zero after you do the ground shrimp procedure, and then you could just simply discard them. Yes. So then, very nice thing about about the fact that we end up with orthogonal Vector&#x27;s is that provided that they&#x27;re non zero. You can be guaranteed that the role in nearly independence. And, uh, yeah, as an exercise tried to prove that fact. Try to prove that orthogonal vector&#x27;s island nearly independence. Okay, um, so this is this question done. Okay, but I want to give a quick glimpse into how I think about the ground shrimp procedure when you have three vectors, because it might get a little bit confusing us too. Why we do all of this? So this is the intuition that I haven&#x27;t least let&#x27;s start off with no three vectors that are that spanned some subspace w. Okay, so, as always, I call X one V one not to produce V two. I simply make X two orthogonal to V one. I don&#x27;t care about X three or any other vectors that I&#x27;ve got floating around. So to make this vector orthogonal tau V one well, I&#x27;m just going to do as we&#x27;ve done always before this. Right? So this makes it This makes me too perpendicular to be one So what happens is that this x two slides over and becomes perpendicular to be one. Now, call that be to and it stays on the plane that spanned by x two and be one. Okay, so now look, the picture of the one is perpendicular to view B two b one a perpendicular, but x three might not be perpendicular to any of them. Okay, so this is the bit where we have to start thinking. So let&#x27;s say that when I say that it kind of goes off like this extreme. So how can I make she let me be consistent to what I had already. So how can I make extremely perpendicular to B one and B two? Well, let me first make it perpendicular to be one. So I define X three dash to be extreme, minus extra dot ve won over a squared comes we want can That gets me too Well, I end up staying on the plan on the plane. Sorry, spanned by these two vectors. Okay. And I end up perpendicular to be one. Subtract by the projection. No, the thing is is that this extra dash might not be perpendicular to v two kind of looks like it is, but it could be there could be, you know, this could be at an angle. Right? So how can I get extra dash to be perpendicular to be, too, while still being perpendicular to be one? Well, the answer is that I take extra dash and I subtract it. Bye. Well, extra Dodge Dart V two. Could you spare time to be too, right? So what I do is I slide this vector over, make it perpendicular TV too. Okay, so let me call this x three. Double dash on dhe. When I do that, I stay in the plane. Spanned by V two the next three. Right? So I stay within this plane. Okay. Now, here&#x27;s the really nice thing. So x three double dash is perpendicular to be too, right. But since but But because extra dashes perp addicted to V one, then this entire plane hostile perpendicular to be one right, because X three dash lives in it. And so since I never leave this plane spanned by V to an extra dashed, I could be guaranteed that my new vector will also still be perpendicular to be one Okay, so it all comes down to the fact that, you know, I found extra dash. And so therefore, I&#x27;ve produced the plane that is perp addictive. Eat, be one. So all I have left to do is to make this guy perpendicular to be too. And so this is V three. So the three is nothing but this and now substitutes this guy into here and you&#x27;ll find that that&#x27;s extreme minus extra dot ve won over what we want squared be one and the minus extra dot ve to over monthly to square turns me too. So that&#x27;s really messy. Apologized. But basically, if you want to find the piece of actor some sequence, then it&#x27;s just x p minus the sum of x p dot the I over Morphy. I squared 23 I and that is the more general formula

Pagadala Kishore Reddy · Answer

we&#x27;re going. I didn&#x27;t Yeah. Do you? Thank you. You okay?

Chris Trentman · Answer

were given a matrix and were asked to determine if the Matrix is orthogonal and, if so, to find the Matrix&#x27;s inverse matrix. A is 2/3 2 3rd 1/3 0 1/3 negative 2/3 5/3 negative, 4/3 negative 2/3 to determine if a is an orthogonal matrix. Get to determine if the column and row vectors of a are Ortho normal to each other. So we have that 2/3 zero 5/3 started with 2/3. 1/3 native 4/3 is equal to well, this is going before nineths plus zero minus 29th is equal to negative 16 9th which is clearly not equal to zero. So it follows that these vectors are not orthe ogle, and therefore it follows that a is not an orthogonal matrix.

Pagadala Kishore Reddy · Answer

it&#x27;s given that dramatic C equals two. One by three Toe by three by three two by 31 My three minus two by 22 by three Minus two by 31 by three No a transpose Sequins one by three two by three Kobe three then two by three one by three minus to wait three and the hard colonies two by three minus two by three, one by a tree. If you want to play with magic c eight mitt transpose you will Yet one gentle, gentle Gino one Judo geul Geul Want the cheese equal to Okay, so we can say that the Grand Matics is Hey, our doe Gordon Magics. And we know that in tow inwards equals two. Right, So here a transfer was equals two The n word inverse matrix I So the investments metrics Amos equals two on my 32 by three to wait three to bay three one by three minus two by three To weigh three minus two by 31 break. Thank you for watching

Pagadala Kishore Reddy · Answer

let my taxi calls toe 11 one minus one. No, the eight clans spores equals 11 one minus one now in tow. Hit on. Suppose he quarrels. One plus one one minus one. The third element is one minus one Foretell mentees. One place for that equals two to Jiro General, go into a trance Particle stones to 00 toe which is not equal to toe identity. Matics I So we can say that given Matics here is not here. Are you gonna metrics? Thank you for watching.

Problem

In Exercises 13 and $14,$ the columns of $Q$ were…

Problem 12 Medium Difficulty

Find an orthogonal basis for the column space of each matrix in Exercises $9-12$ .
$$
\left[\begin{array}{rrr}{1} & {3} & {5} \\ {-1} & {-3} & {1} \\ {0} & {2} & {3} \\ {1} & {5} & {2} \\ {1} & {5} & {8}\end{array}\right]
$$

Answer

$\left\{ \left[ \begin{array} { c } { 1 } \\ { - 1 } \\ { 0 } \\ { 1 } \\ { 1 } \end{array} \right] , \left[ \begin{array} { c } { - 1 } \\ { 1 } \\ { 2 } \\ { 1 } \\ { 1 } \end{array} \right] \cdot \left[ \begin{array} { c } { 3 } \\ { 3 } \\ { 0 } \\ { - 3 } \\ { 3 } \end{array} \right] \right\}$

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

Linear Algebra and Its Applications

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Caleb E.

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Vectors Intro

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$\left\{ \left[ \begin{array} { c } { 1 } \\ { - 1 } \\ { 0 } \\ { 1 } \\ { 1 } \end{array} \right] , \left[ \begin{array} { c } { - 1 } \\ { 1 } \\ { 2 } \\ { 1 } \\ { 1 } \end{array} \right] \cdot \left[ \begin{array} { c } { 3 } \\ { 3 } \\ { 0 } \\ { - 3 } \\ { 3 } \end{array} \right] \right\}$

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Caleb E.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Anna Marie V.

Heather Z.

Caleb E.

Michael J.

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

Linear Algebra and Its Applications