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Prove Theorem 3 as follows: Given an $m \times n$ matrix $A,$ an element in $\operatorname{Col} A$ has the form $A \mathbf{x}$ for some $\mathbf{x}$ in $\mathbb{R}^{n} .$ Let $A \mathbf{x}$ and $A \mathbf{w}$ represent any two vectors in $\operatorname{Col} A$ a. Explain why the zero vector is in Col $A$ .b. Show that the vector $A \mathbf{x}+A \mathbf{w}$ is in Col $A$c. Given a scalar $c,$ show that $c(A \mathbf{x})$ is in $\operatorname{Col} A$

a. the zero vector is in $\operatorname { Col } A$b. $A x + A w \in \operatorname { Col } A$c. $c ( A x ) \in \operatorname { Col } A$

Calculus 3

Chapter 4

Vector Spaces

Section 2

Null Spaces, Column Spaces, and Linear Transformations

Vectors

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

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in this video we start off with a matrix A which is going to be of size and buy in. And what our goal here is to do is prove that the calm space of a is a subspace of RM. Where the m came from the first entry. Well, there is a really fast proof for this. Let's show that proof first. So the calm space of a we know by definition is thespian man of the columns of a while. The columns have em entries because it's an M by N matrix. We already know it's a subset of RM, but theorem, one of chapter four is a very, very helpful here. Um, it tells us immediately that this is a subspace and so the proof is complete. Unfortunately, we're going to do this in a slightly harder way. But it's good to know just how powerful here and one is a soon as we have a span, we know we have a subspace. Let's go back to this problem here, and we're going to prove that the calm space of a is a subspace of RM by using the definition of subspace. So what? This problem is Maura about than anything is just getting used to that definition of subspace. So there's three parts in this approach in the first part? Well, no, we really have a subspace. If we can first show zero Vector is in the calm space. Let's write out the calm space in a different way to help us with this proof. We know that the calm space of a can be expressed as the set of all vectors be such that B is the same as a Times X for some X in our, um, so that's what the calm spaces we need to show that the zero vector is in this set up I meant to highlight is in this set. So start off with a column vector or the zero vector will say, Let zero and I'll write it as zero sub and by one So we know it's the zero factor off our M. So we'll say, Let that be the zero vector of RM. Now the defining equation that gets us inside the calm space is this right here? We're putting the zero vector in this place, so that's what we're going to start with. I'll say here then the zero vector of size and buy one is going to be equal to a matrix, a times a particular vector X, which must come from our end. Let's take for our vector X another zero vector of size and buy one. We don't that a matrix eight times of zero vector definitely gives us. There's your vector back. And so this equation is true. And then that implies immediately that the zero vector of size and buy one is in the calm space of a. So that's our first step in proving we have a subspace. By definition, the second step is to do the following start by saying, Let let's say B one and be to be in the calm space of a So our goal here is the following. We want to show B one plus B two is in calm space of A If we can show that B one plus B two is in the calm space than our work will be complete. So how do we do that? Well, first, let's start with the assumption that we made B one B two are in the calm space and using this way of visualizing the calm space to guide us. It tells us that there exists x one and x two. It are in such that if we take eight times, X one will get B one and eight times X two. Must be be to. These statements are valid only because we said B one and B two came from the calm space. And so it must satisfy that equation. But recall our goal is to look at the addition of these two vectors. So let's do that next weaken, say then B one plus B two is equal to well, first in place of the one we have a X one plus for B two, we have a X two, and our goal is to write this as a matrix. A times a particular vector. X Well, we have a mess here then, but we do have a strategy we can left factor a and leave x one plus x two. Then this is our vector X. Let's put it in blue for emphasis. So we just showed that if we take B one plus b two, we have a vector x one plus x two times a giving us B one plus B two, then that immediately implies that B one plus B two is in the calm space of a So step two is now complete. This takes us to the third and final step. We say this time let be being calm space of a And this time we're also using a scaler. See, and let's let CB in are just to be specific What see is and where it comes from. Well, because we said be is in the column space. We can now say that there exists X in our n such that eight times X is equal to be. And let's also state what our goal here is. We need to show that C Times be is in the calm space of a So we're going to start off the left hand side here on the definition. But use C times be well. We know by substitution that B is equal to eight times X. It's all place that here and once again, our goal is to find or to write our answer in this arrangement eight times a particular vector X So going back down here, we don't have quite that form, but the scaler c can commute as follows. We can say that there's a matrix X which is multiplying. See Time's X Also see it just for the sake of emphasis. Well, that's our vector. And so this implies immediately that C B is in the column space of A. So our work is not quite complete. Recall that we need to show all conclusions and we now have that since 12 and three are satisfied in the definition of subspace, our conclusion is as follows. Let's right the conclusion and purple. I will say this shows that the calm space of a is a subspace of RM as required. So this is the proof by using the definition of subspace. But this is again a great time to plug that amazing theorem One. If you're allowed to use theorem one, then this is the proof as opposed to what we just did over here.

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