suppose-aq-r-where-r-is-an-invertible-matrix-show-that-a-and-q-have-the-same-column-space-ihint-give

Name: suppose aq r where r is an invertible matrix show that a and q have the same column space ihint give
Uploaded: 2019-09-01
Duration: 10 min 11 s
Channel: Numerade
Description: Suppose $A=Q R,$ where $R$ is an invertible matrix. Show that $A$ and $Q$ have the same column space. IHint: Given $y$ in Col $A,$ show that $\mathbf{y}=Q \mathbf{x}$ for some $\mathbf{x} .$ Also, given $\mathbf{y}$ in $\operatorname{Col} Q$

Question

Suppose $A=Q R,$ where $R$ is an invertible matrix. Show that $A$ and $Q$ have the same column space. IHint: Given $y$ in Col $A,$ show that $\mathbf{y}=Q \mathbf{x}$ for some $\mathbf{x} .$ Also, given $\mathbf{y}$ in $\operatorname{Col} Q$

Sean Thrasher · Accepted Answer

given this problem where a is a cute times are where are some inverted matrix? We need to show that a and you have the same column space. So as sets, um, we need to show that called a is equal to call Cute. Now a useful method of proven maths is to show. Well, if I want to show the two sets are equal, I must show that Diskant is contained in this set on the other way around as well. Okay, so if you like a Zeke to be where I am BR sets if and only if days including and B on B is included in a Okay, All right, so this allows us to tackle this problem bit by bit. So let&#x27;s first show that the column space of a is contained in the column space of Q. Well, if you want to show that one set is contained in another. So let&#x27;s say if you want to show that exit contained, and why show that&#x27;s if you pick any little X from this set and that must mean mats. Same. The Lexx is and why, right? You know, if X is contained in why than any point here is automatically in. This said that contains it. Okay. All right. So what does this mean? So we must show that if you take why in the column space of a That means that why isn&#x27;t the column space of cute? All right, so what does that bull down too? Well, this is the very last step that we want to arrive at and a useful method. Know yet another method, Another tactic and proving things and maths. Just to think what will be one step before my conclusion, What will be the penultimate step? Well, if I&#x27;m gonna show that, why is in the column space of Q? I have to show that&#x27;s there exists some U in our in such that why is equal to Q times. You okay? Why is that? Well, if you go to page 203 and look at the&#x27;re, um three, that pretty much explains it, but let me really quickly go over it, because it might help. Well, you know. So you have to be careful. This statement right here will imply this. And so if I show that why being the column space of a implies this you know, just knowing that you&#x27;re in the column space of a you&#x27;re guaranteed. You know that Y is equal to Q times you. Then that&#x27;s enough to conclude that. Why is in the column space of cute? Hey, so why is it that this statement is true? Well, OK, pause and try to figure this out yourself because it&#x27;s actually not that bad. Okay, so let&#x27;s see that Q was built up off the common factors. Q. One up to Q m and then you is the column vector you up to you? And where this These are the entries and the four arrows on here. Well, Q times you is the same thing as you. One times Q. One up to U. N Times Q. Ed and hey, I&#x27;ve got a linear combination of the column vectors and so cute times you, which is just why has to sit in my column space. So that&#x27;s the reason behind Theorem three on page 203. Okay, so how can we actually proved that? Let&#x27;s have a think. So how can I improve this fact? Okay, we&#x27;ll say that&#x27;s supposed, But why is in the column space of a then well, go to your text book again, and I want you to look at page 206. It says from the little from that box on the top, I think it&#x27;s an under bullet bullet 2060.5 that why? Being in the column space of a implies that the equation eight times X people toe Why is consistent now? What do they mean by this? Well, all they mean is that if why is in the column space of A that implies that there exists some X in our in such that y is equal to eight times X. That&#x27;s all they mean. You can solve this equation for X. It exists ex exists. You know, it could be the zero vector, or it could be anything else, but it exists. And so clearly, you know, you can see you might see the connection between these two already, But if you don&#x27;t let me go through it, Okay, so we&#x27;ve got that. Why in the column space of a implies that there exists some X and rn such that y z eight times X. But ese click, you times are right. So that means that there exists an X in Oran Such that y is equal to Q times are times X But now our times x you know, let me call it you. That&#x27;s just another vector in our and OK And so we found that. Why being the columns base of a means that there exists a mu in rn where I&#x27;ve actually told you what this you is So you can be confident that exists such that Why&#x27;s it click you times you and we&#x27;re done showing this way around. Okay, so if you&#x27;re confused by this, let me go over this one more time. So I want to show that column Space of a is contained in the column space of Q. So I took why in the common space of a and I said Okay, that means that there exists some um, so there exists an X in are in such that wise it eight times x. But then, you know hey, is just Q times are I have to use that fact right on dso letting you so there exists a mu which is equal to our times X in our in such That&#x27;s why is it the Q Times you and therefore, why is in the column base of Cute? All right, so this was what I proved boiled down to. Now, let&#x27;s go the other way around. I want you to pause this on and try this out yourself. You should be able to do this if you can&#x27;t go back and watch the video again. Okay. So we want to show, you know, it&#x27;s very, very similar now. And by the way, this method actually was given in the hints. Okay. No. Suppose that wise in the common space of Q, I want to show, but Roy is in the columns base of a All right? So that means that I want to find some you and are in such that such that. What such that? Why is equal to a times You all right. So this is we keep this in the back of our minds. All right? You know, this might help guide us with what facts we choose. So again, as I said before, Why being in the common space of you simply means what? That there exists some X in our in such that y z Q times X, no. I need to bring a back into this. So a Z Q times are right now. Our is in vertebral, so I could multiply on the right hand side by our in burst. Okay, okay. And so this guy right here becomes my identity. Q times the identities just cute. So accusing eight times are in verse. Okay, no deal around. Queues no are inverse. Times ain&#x27;t necessarily That&#x27;s what&#x27;s so I had to use the fact that our is in vertical That&#x27;s important. Otherwise it would not have necessarily been true. Okay, so they exist in X and rn such that y is equal to Q times X. So that means that their existence are in such that y is equal to a are inverse times X But our again, that&#x27;s just some other transformation from our in tow or in. And so are inverse times x. You know, let me call that you That is some vector in rn. So therefore, you know it&#x27;s not the same is the you that I had before? But who cares? Oh, so therefore, we conclude that there exists a Mu and Orrin such that Why is it eight times you and we&#x27;re done? So we&#x27;ve proven this way around. We&#x27;ve proven this way around And so we proven this on with Dr

Angelo Rendina · Answer

So we have a times the is equal to the identity metrics. We want to show that the is the inverse of a by using metrics algebra. So since a is in veritable, we can multiply both sides from the left by the inverse of a So we have a minus one a d equal a minus one identity. Now here we use the associative property of matrix multiplication toe put the brackets between the first and second factor So a minus one, eh? And only after we multiply by d and now the right that side multiply my identity Medics does nothing. So we&#x27;re just left with a minus one Now, by definition, a minus one times a is exactly the identity matrix. So these old product here is just identity that multiplies the equal to a minus one And again we&#x27;re deploying by idiot entity medics there&#x27;s nothing. So on the left hand side we have the and then equal a minus one, which is what we wanted

Jimmy Yao · Answer

way need to apply. Just urine. Four Queen 9.1. So it is the ring, um, to rink nal ity serum. So it stays for any A is m by and the rank of A Because the minority of a it&#x27;s the culture in into this province It gave up its condition. That&#x27;s the common space. So that there Colin Space, it&#x27;s you co two than now space So they aren&#x27;t you go. So if there you go one thing is the dimension of the Colin space this Nico to the dimension of the now space. Right? So the dimension of the constipated should go to the rank of A into by definition, the dimension of the now space. It&#x27;s really equal to the reality that a so here this end of this are the same number. Say, if this is X and this will be X, then end will be two x considered the Dimension X. Here is there is the interchange. So again, it&#x27;s two times I muted. So Ben is even en If even the any it even means there Colin Number is even so a have even college

Angelo Rendina · Answer

so we have be minus C that multiplies the equal to zero. So since the is in vertebral, we can multiply from the right by the inverse of the so right B minus c the and then the inverse. And that&#x27;s equal to zero times Demers. Now the product the D in verse is just it. Entity metrics. So this becomes B minus c that multiplies the again. Did he is equal to now Any metrics multiplied by 00 So on the right hand side, we just have zero and now must be bribed by the identity matrix does nothing. So on the left hand side we have B minus c and on the right inside we have zero Now we some ah si on both sides. So that&#x27;s right, B minus c and then we add plus C on the right. Inside we have zero Blasi Well now minus C plus C is just zero. So are we on the left inside? We just have B and on the right inside something zero does nothing. So we&#x27;re just left with C, which is what we wanted

Joshua Eastwood · Answer

this question tells us that B squared for some Matrix B is equal to zero and we&#x27;re trying to show that I minus b is in vertebral by showing the inverse is I pop be okay, So we want to show that the invert my mind if we know that he squared is equal to zero So here&#x27;s what we&#x27;re gonna do We&#x27;re gonna take I minus b and multiply it by I want to be And if we do that, we get I weird I&#x27;m done Plus, um I times b and then mine It&#x27;s b times I and then minus Pete spared is equal to music row And that just gives the I there is I These are going t minus B is gonna be nothing. Hey, that&#x27;s making me zero. And he squared, we know is zero. So I minus b times which is what I wanted to show because I wanted to, but I might is the invert I i plus b is the inverse of my mind. So if I then do the other way around I plus b multiplied by I minus B because I get it the only thing that different get mine I d in a plus B, I Okay, but again, everything goes to zero. Has shown that this is true because by minus be kind, I must biggest get any major picked. I thought me Yes. Oh, is the inverse of this one?

Griffin Johnston · Answer

suppose I had a square matrix. A So a is an in by in matric it has the same number of rows as column. And suppose also that the road space of a is equal to the null space of a no space of a And just as a quick reminder than no space of a is all the vectors x the set of all the vectors x such that a times the vector X is equal to zero. It&#x27;s all solutions to the homogeneous equation with this matrix. No, we want to prove in this video that a prove a the A cannot be convertible now, like I talked about in previous videos Ah, good first initial investigation for proof problems especially like this one. We&#x27;re trying to show that a or some some some mathematical object in this case AIDS and matrix. We&#x27;re trying to show that the A cannot have a certain property. In this case, we&#x27;re trying to show that they cannot be in vertebral, usually a good first initial approach to kind of get your hands dirty, and sometimes it ends up being the proof. It ends up turning into a foolproof, but regardless a good initial approaches to attempt to prove this by contradiction. So what is our goal? We try to prove something by contradiction. Well, we first suppose the contrary to suppose suppose a can be convertible can be convertible, then our goal, our goal when we suppose this sister derive a chain off implications. So if a driver change implications, what I mean by that is supposed taken, be in verbal without implies something. And that that&#x27;s something implies another something, and so on and so forth and to eventually we this the end of this chain of implications. We arrived at the end of his chain of implications. We arrive at some contradiction and maybe not the end, but one of these statements in here in this chain of contradiction, sometimes it&#x27;s we we stop when we arrive at a contradiction, so it ends up being the end of our chain. But it&#x27;s not necessarily the end of the Cheney. This infiltrated implications could go on forever, but regardless we eventually arrive at some contradiction. And when we arrive at a contradiction, that means that if we were supposed, if we were to suppose a can be in vertebral, then that would imply a contradiction. It would imply a statement that is absurd. That contradicts our original assumptions Are original assumptions that were given to us to be true for this particular problem. And thus this statement right here Supposing a can be in vertebral that can&#x27;t possibly drew a cannot be in vertebral if you were to arrive at a contradiction. Therefore, A Like I says, I said a cannot be convertible and we&#x27;ve proven a statement. So think for a minute what would happen if a is in verbal and make sure we want to leverage the fact that the road space of a in the north face of a are equal. So what would happen if a is and vertebral and we also have that the roast base of a is equal to the no space of a So I&#x27;m assuming you&#x27;ve had a go at it. So when a is in vertebral by the in vertebral matrix here, Um, all right, the in vertebral matrix here, um, there&#x27;s two things that we know. One of the things that we know if that if an matrix is in verbal, then the Onley solution x to the homogeneous equation A time of the vector X is equal to the zero vector. The only solution is the trivial solution. So X equals zero Vector is the Onley solution to this equation the Onley solution to this equation the Onley solution to this equation. But if X is equal to this year vector that that&#x27;s the only solution to this equation. Remember the set of all acts such that access solution to this equation is the knoll space of a So that means that the north face of a the north face of a is equal to the set containing zero vector and that&#x27;s it. That&#x27;s the only vector in the north place of a now something else that thean vertebral matrix theorem says is that if I&#x27;m matrixes in vertebral, which we are supposing a to b in vertical, remember by our by our our attempted a proof of contradiction proof by contradiction, we&#x27;re supposing that a is in vertebral. So if they were to be in vertebral then this will be true and also by the in vertical Matrix syndrome A is row equivalent to the end by in identity matrix And check out what we know we know that if two matrices are row equivalent, then that means they&#x27;re row spaces are equal. So the road space of a is equal, since a is row equivalent to the identity matrix because a convertible by the in vertebral matrix serum again, we&#x27;re supposing it to be in vertebral here, just just to see what happens to see if we arrive at a contradiction. So since we&#x27;re supposing it to be in vertebral, we know by the invariable major theme it&#x27;s row equivalent to the identity matrix. And we know there&#x27;s a theory that says that if two matrices aero equivalent, then that means that there rose spaces are the same. So we can say that the Roadway survey is the same as the rose base of the identity matrix, but the road space of the identity matrix, the in by an identity matrix. I&#x27;ll just write a kind of a scratch example of the identity matrix, so each vector looks a little bit like this, and finally we get to the last vector. In this I didn t matrix the identity matrix, the road space of the identity makers. It&#x27;s the span of all the row vectors now the span of all these row vectors is all of our in and namely this family&#x27;s row vectors for sure has the vector 1000 and so on, all the way to in or in minus one number of zeros. So the span of all these row vectors is definitely not equal to the set containing Onley, the viewer vector. Moreover, the span of all these row vectors actually equals all of our in So So let me let me emphasize this year. So So this span of all these reflectors is all in your combinations of these row vectors. So, for instance, one vector that would be in the row space off a would be the vector 100 all the way up two in minus one zeros bluff 01 all the way up two in minus one zeros, including this first year right here. And this vector is equal to 11 and then zeros for the rest of the entries. And this vector right here the fact that this factor is in the road space of a because any linear combination of the row vectors off the identity matrix, any linear combination of the row vectors. Identity matrix will lie in the span of the row vectors, a k a in the rose base off the identity matrix, which is the same as the space of a So therefore, the fact that this vector the linear combination of the other of the row vectors in this matrix, these two vectors plus zero times all the other vectors this vector lies in the road space of a. But remember, we suppose that the road space of a is equal to the no space of a. That means if the roads base of a sequel to the North based pay we were given this to be true. That means that every vector on the road space of Asia is in the north face of a and every vector in the north face of a is also in the rose face of a. But check this out. I found ah, vector in the road space of a That is definitely not in the north face of a. So the fact that a is in vertebral implies a contradiction, namely, the fact that we found a vector in the road space of a that is not in the north face of a implying that the road space of A is not equal to the North Bay survey which contradicts what we were given to be true. So if we were to suppose that a can be in vertebral, we would arrive at a contradiction at a contradiction. So therefore a cannot be in vertebral and we&#x27;re done.

Problem

Given $A=Q R$ as in Theorem $12,$ describe how to…

Problem 20 Hard Difficulty

Suppose $A=Q R,$ where $R$ is an invertible matrix. Show that $A$ and $Q$ have the same column space. IHint: Given $y$ in Col $A,$ show that $\mathbf{y}=Q \mathbf{x}$ for some $\mathbf{x} .$ Also, given $\mathbf{y}$ in $\operatorname{Col} Q$

Answer

See proof.

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

Linear Algebra and Its Applications

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

Catherine R.

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

Linear Algebra and Its Applications