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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 $\operatorname{Col} Q$

See proof.

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 4

The Gram–Schmidt Process

Vectors

Missouri State University

Campbell University

Harvey Mudd College

Baylor University

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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'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's say if you want to show that exit contained, and why show that'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'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'm gonna show that, why is in the column space of Q? I have to show that'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'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're in the column space of a you're guaranteed. You know that Y is equal to Q times you. Then that'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's actually not that bad. Okay, so let'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'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's the reason behind Theorem three on page 203. Okay, so how can we actually proved that? Let's have a think. So how can I improve this fact? Okay, we'll say that'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'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'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't let me go through it, Okay, so we'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'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've actually told you what this you is So you can be confident that exists such that Why's it click you times you and we're done showing this way around. Okay, so if you'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'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'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't go back and watch the video again. Okay. So we want to show, you know, it'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't necessarily That's what's so I had to use the fact that our is in vertical That'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'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'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're done? So we've proven this way around. We've proven this way around And so we proven this on with Dr

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