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In Exercises 21 and $22,$ mark each statement True or False. Justify each answer.a. A subset $H$ of $\mathbb{R}^{n}$ is a subspace if the zero vector is in $H .$b. Given vectors $\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}$ in $\mathbb{R}^{n},$ the set of all linear combinations of these vectors is a subspace of $\mathbb{R}^{n} .$c. The null space of an $m \times n$ matrix is a subspace of $\mathbb{R}^{n}$ .d. The column space of a matrix $A$ is the set of solutions of $A \mathbf{x}=\mathbf{b}$e. If $B$ is an echelon form of a matrix $A$ , then the pivot columns of $B$ form a basis for $\operatorname{Col} A .$

a. Falseb. Truec. Trued. Falsee. False

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Gideon I.

Algebra

Chapter 2

Matrix Algebra

Section 8

Subspaces of Rn

Introduction to Matrices

Missouri State University

Oregon State University

Harvey Mudd College

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right. So they want us to determine if these statements are going to be true or falls. Um, so this first one, it says essentially, that we have the zero vector and H. So this is a subspace of our in and this is going to be false because it doesn't say anything about the other two properties that we need to have, which is it Closed under sums and closed under scaler multiplication. So false. Because we do not No. If actually, let me beat the Patriots. Bigger. Do not know if closed under some and scaler multiplication. Okay, so we have a there now for B. They say we have this set of factors, and the set of all linear combinations is a subspace. And so this one is going to be true, because if we just think about, um, what this is so it would be something like see one V one plus c two, b, two plus dot, dot dot all the way up to C p v p. So we can get the, um zero vector by saying all see, I equal to zero for every eye. So that's fine. Um, we can get any scaler out of this by saying, see, hi equals zero except for some value. Um, but I is equal to and where it is, like the vector that we want. So that gives us the scale of property and three, some wolf. I mean, some just kind of comes from the definition. So some is, um, true by definition of linear combination linear combo. So that's how we end up getting true for B. See, I believe this one is also true, but let's go ahead and kind of draw it out. So if we have over here some Mbai in Matrix, remember what the knoll space really is? It's saying, Well, what collections of vectors here output the zero vector. And, um, in order for us to multiply these we needed in by one matrix, which means we would have in entries here and then this is an element of our in. So then the null space has to be a subspace of our because it will have the zero vector on. We'll also have where if we add two things by the linearity of matrix, that should be fine. And then if we just multiply this, that should also be fine. Um, well, actually, let's go ahead and write all those out. So eso let's say just so we can kind of be a little bit mawr exact right? So Ah, we have some matrix A with the vectors. Let's just call it X Y elements of the knoll of a All right. So if we want to do the some property So let's look at a Times X plus y Well, then this is going to be a X plus a y and by definition of both of these being in the null space, this is the zero vector plus zero vector which is zero vector. And so then that implies X Plus y is also an element of nor a Okay, so we have that one and then first scale er's. So this is some. So that's good. I guess we should have also did the zero vector. But that one's kind of true just by definition. So this is zero element of Norway. Alright, so that's good so far and then lastly we need these scaler. So we have a times c of X, remember axes so in the north space. But we can go ahead. And do you see? Actually me, Right This over here. So this is going to be C a X like that so we can apply community since he is just some constant. And then we'll that c times zero vector, which is equal to zero vector. So then that implies C X is also in the knoll. So then the constant multiple or the scaler also holds. So, um, I don't know if you really need to be that exact, but, I mean, it doesn't really hurt now for D the column space of Matrix A is the set of solutions. A X is equal to be all right. And so this is actually going to be, uh, false. And the reason why this is false is because it's not necessarily just for the solution of one. It's supposed to be the span of all of the columns I'm sorry of. Ah, yeah, the span of all of the columns. Um, so, yeah, that's why this one isn't necessarily true. Because it's not talking about a solution set and then for E. So this is saying that B is the echelon form of a matrix A in the pivot columns of B form a basis for a And so this one, um, is also false Because it's not the pivots of B, but the pivots of call a Arab the pivots of a instead. So this should actually say a here on beacon, actually kind of show that off on the side really fast. Eso Let's come down here and do that, right? So let's say we have some matrix here. And so this is kind of like you can use, like, a counter example as to why this wouldn't be the case. Um, so we have, like, 123 uh, then 567 13, 14, 15. And then, I don't know, negative 112 So hopefully if I reproduce this year Ah, this gives me 111 000 And if it doesn't just kind of tweak this so it does. But I believe this will give us that. Okay, well, if we look at it, this is a and then this is the echelon form. Be over here. Well, if we were to look at this, though, over here, this is saying if we take each of these are fourth position of our vector is always going to be zero. But over here, notice There are numbers here. So we wouldn't even be able to get out the, like, original columns here. So because of that, um, we would need to use the columns for a as opposed to the columns of being So Yeah. So that's kind of like the rationale behind. Why that? Why that ISS? Yeah. So s it looks like the first statements. False second statement. True, They're true. And then the last two are going to be false.

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