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Question

Let $H$ be the set of all vectors of the form $\left[\begin{array}{c}{s} \ {3 s} \ {2 s}\end{array}ight] .$ Find a vector $\mathbf{v}$ in $\mathbb{R}^{3}$ such that $H=\operatorname{Span}\{\mathbf{v}\} .$ Why does this show that $H$ is a subspace of $\mathbb{R}^{3} ?$

Daniel Pezzi · Accepted Answer

Hello. For this problem, we need to show that we can write this vector as a if I call the set H, which is just the set of all vectors of this form or s is some arbitrary element of our We need to show that we can write H as this span of some vector the which will also show that h is a subspace of our three. The reason why this shows that h is a subspace of our three comes from the definition off span. If you if you remember the conditions that a set has to meet to be a subspace if if If all of the elements are members of a larger vector space, those air exactly the conditions is there exactly what defines a span of the of a given vector. So this one is relatively simple because remember, the span of a single vector is just all arbitrary constants s in our field. In this case, the real numbers times that vector. If we look here, we can immediately just factor out in s and get that h is equal toe s times. The vector one, 32 again s is just some arbitrary, constant, so h is equal to the span of our vector. 132 The notation gets a little bit sloppy when you&#x27;re actually doing with real vectors. But this just shows that this is just a non arbitrary, constant times this given vector on we&#x27;re done.

Daniel Pezzi · Answer

Hello. For this question, we&#x27;re giving a subset of our three, which is the set of all vectors of this form where B and C and R are arbitrary constants in our field are and we want to write w as this span of two vectors. Now it&#x27;s important to remember the definition of span is just the set of all vectors that could be written. I&#x27;m gonna cheat a little bit looking at this and use the same constance of be times you plus C times V, which is the same thing as the set of all linear combinations of the vectors inside these brackets. So, looking at this, it becomes obvious what we need to dio. We need to break up this combined vector into two separate vectors multiplied by constant B and the constant See. So I Kenbrell up this over addition. So I&#x27;m going to have this is what I want to end up with, and I want to fill in what these vectors are. So I have a five multiplied by a B here, my second slot, I have one multiplied times to be, and then in my third slot, I have nothing multiplied times to be. So I leave a zero here and then when I look at this factor, I have a to multiply times to see. I have nothing multiplied times to see here. And then I have a one multiplied by a c here. And if you don&#x27;t believe this process, you could just multiply through and add the two together and you&#x27;ll end up back at this spot. But now that we&#x27;ve rewritten this vector in an equivalent form, it becomes immediately obvious. What are two vectors? Are you or or V? It doesn&#x27;t matter which you which one you denote is our vector. Call this you 510 and this vector here is R. V 201 and now we have written W as the span of these two factors here and

Viswesh Uppalapati · Answer

and this problem, we&#x27;re gonna be solving problem number 10 of section 4.4. Um, so the problem assets, if the given set are all vectors of this given definition, Where to? Where? We have a column vector of two teas there on negative to t. Here&#x27;s some space of our three, which basically just means that, um, it spans are three. So for a given set, to be a subspace of another set, we need to fulfil three conditions. One it has to contain the zero vector to it has to be close under addition and three, it has be closed on her multiplication. So, um, if we just take a vector V at its base form to zero negative to t zero and negative t and we take t to Tia&#x27;s any scaler values. So if we take t to be zero, we got two times 00 negative zero equal, which is 000 which is the zero vector. So the zero vector exists in this set so we can go on to the second condition. Clothes under a scaler are closing tradition. So if you take two vectors in this set where they multiply by a different scale our values So we can have V be to t zero negative t and you be to see zero negative c and we add these two together we get V plus you times are sorry. It would just be replace you. Oh, sorry. We Plus you would equal to to t plus to see you&#x27;re a plus zero and negative t minus Siebel That is just the same thing as adding adding the scaler multiples together t plus c times this vector 201 which are just 20 negative one which is just the coefficients of the initial vectors. So this means that the vector remains the same but the scale of multiple just changes. It just adds and compounds. That means that we&#x27;re still on the same line with varying scaler values and we&#x27;re still within the same set. So it has closed under Skeletor, Earth is closed under addition and lastly, it has to be closed in her scaler multiplication. So if you just do, um, if you just do a scaler scene times the vector V, we got see Time&#x27;s see Time&#x27;s to t zero negative t That is just C Times, T times two 20 negative one Which again, is still this the same coefficients of the set? Um, So we still stay on the same line again? Because I just just the scale of all you were multiplying this by changes. See, times to you just get bigger. But we&#x27;re still scaling the same line or vector. So we stay on the same line. Um, therefore, this is also closed on her skin. Her multiplication therefore, um h is a so space is a subsidies o r three

Viswesh Uppalapati · Answer

in this video. I&#x27;m selling problem number 12 of section for my one. And, um basically, it asked us a show. It, uh, got said w is a subspace of are for, um, and said, w contains all the vectors in the form that fit this definition right here. I suppose three. T s minus two U two s minus t and for us for tea. Ah, as like the values of the vector. So if we set, don&#x27;t you said w to be span UV? We know that, um, you as 11 to 0 because it&#x27;s just the coefficients of the S terms in this definition, and V is three negative or negative one for because it&#x27;s the coefficient off the tee terms in this definition that was given to us. So, um, we know that w must w set that spans u V with just means that it contains all the doctors in the in the form of UV added together like like this, given by the scientific definition. So to determine if probably use a subspace of, um, are for we have to make sure that it fits these three conditions. The definition of W W fits these conditions. One is that it must contain the zero vector. Um, and two is that it has to be closed under addition, And three is that it has to be closed in her skill in multiplication. Instead of trying to prove every single one of these, there is the easier way to solve this foam where theorem one just says that, um, someone from the section just says that if ve won t v p are in a better space V then span the one through three p is a subspace off the So here we find w span you envy already. And we know that you envy are in the set W because we&#x27;re given this definition. So by theorem one w is a subspace love are for oh

Donald Albin · Answer

so that the set of vectors and I&#x27;m going to write it down here, um, does not span three dimensional riel space. But it does span the subspace of three dimensional space consisting of all vectors lying in the plane with equation X minus two y plus Z equals zero. So I&#x27;m going to write down the vectors 345 and, um, it will spend this subset this plane. All right, So if we right that any vector is going to be a constant Times v one plus a constant Times V two plus a constant times V three, I should note that this is V one. This is V two, and this is V three, then a constant times 12 three. That&#x27;s V one constant times 345 Let&#x27;s v two and a constant times that which is V three. If I, um, right down the X coordinates, it would be one times see one plus three times C two plus four times C three is the X coordinate two times. See one plus four times C two plus five times C three is the y coordinate. And just to be clear, that&#x27;s where I&#x27;m getting them from. And to be more clear, this equals X. Why Z any ah vector in three dimensional space. Okay, also three time See one plus five times C two plus six times C three is ze. Now we can write this as an augmented matrix. 13 four, 245 356 I got them from right here. 134 to 4 or five, 356 And then the solution. Now the solution. But the X y Z matrix is just x. Why Z the solution matrix would be C one C two C three. All right, we can ah reduce this matrix. Uhm, I&#x27;m going to take too negative too. Times the first rail and add it to the second row. Oh my goodness. That is not what it says. Okay, Two times the first run out of the second rule. That gives us zero negative. Two times three is negative. Six plus four is negative. Two negative two times four is negative. Eight plus five is negative. Three and ah, negative too. Times X is negative. Two x plus Why? Negative two x plus Why? All right now Gonna take negative three times the first row and added to the third row. So that&#x27;s gonna give me zero negative. Three times three is negative. Nine plus five is negative for negative. Three times four is negative. 12 plus six is negative. Sex negative, three times x plus z Negative three X plus z. All right, let&#x27;s keep working. I&#x27;m going to take negative, too. Times the second row and add it to the third row that&#x27;s going to give me. Didn&#x27;t do anything to the first row. Not doing anything to the second row Third row will be zero negative two times negative to positive. Four plus four is zero negative. Two times negative. Three is positive. Six plus negative six is zero and this is going to be negative two times negative. Two x, which is positive for X negative two times positive. Why is negative? Two. Why minus three acts less Z and from the last row we see that four x minus two. Why my eyes? Three acts plus Z equals zero. So for X minus three X is X. Don&#x27;t do anything with the negative two y plus Z equals zero, and that&#x27;s exactly what we wanted to show X minus two. Why plus Z equals zero. So that set of vectors does not spend the entirety of three dimensional riel space. But it does span a subspace and that subspace that its fans is the Plain X minus two y plus Z equals zero.

Michael Jacobsen · Answer

for this example. We start out with age. It&#x27;s going to be a vector subspace of some vector space V, and then we&#x27;re going to make an assumption that we have two vectors U and V, which are also in H. So what we&#x27;re going to do in this situation is verified the following claims. We claim that the span of those two vectors U and V, is contained in H. I&#x27;ll use a symbol subset to say that the span of UV is a subset of age, which is a cinnamon synonym of saying it&#x27;s contained. Let&#x27;s see how to prove this. First, we begin by taking an element. We could call it X if we like that comes from the span. So let&#x27;s say let X be in the span of the vectors you envy. Our goal here is to show that if X came from the span, then it must be in the set. H if we can show that that our work is done and we would have shown containment. So let&#x27;s go on to the next step. Then, as soon as we know that the Vector X came from this man, that means next that there exists scale er&#x27;s I&#x27;ll call them C and D such that X is equal to see Time&#x27;s You plus d times V. The reason we could write that such an equation like this is FX came from the span of U and V. That just means it&#x27;s a linear combination of you envy. And this is one such linear combination now notice. So far we have nowhere used the fact that H is a vector subspace. So let&#x27;s begin doing that next. We know that since H is a subspace of the and we know that U and V came from H, which is also critically important, it follows that See, you and Devi are in H now why is that true? Well, the reason is, if we have a subspace, a subspace is closed under scaler multiplication. And so this is a scaler multiplication of a vector u a vector v which came from H so they must now both b NH as well. Whatever the closure, do we have? Well, another closure property is if we use vector addition now of two vectors that are known to our day b NH then the some of those vectors must also be in age. I will say likewise X which is equal to see you plus DV is in a church. So this was our goal. This shows that the spin vectors U and V are contained in the subspace H, as required.

Problem

Let $H$ be the set of all vectors of the form $\l…

Problem 9 Easy Difficulty

Let $H$ be the set of all vectors of the form $\left[\begin{array}{c}{s} \\ {3 s} \\ {2 s}\end{array}\right] .$ Find a vector $\mathbf{v}$ in $\mathbb{R}^{3}$ such that $H=\operatorname{Span}\{\mathbf{v}\} .$ Why does this show that $H$ is a subspace of $\mathbb{R}^{3} ?$

Answer

Thus, H is a subspace of $R^{3}$

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

Linear Algebra and Its Applications

Anna Marie V.

Kayleah T.

Caleb E.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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Thus, H is a subspace of $R^{3}$

Linear Algebra and Its Applications

Anna Marie V.

Kayleah T.

Caleb E.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Anna Marie V.

Kayleah T.

Caleb E.

Joseph L.

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

Linear Algebra and Its Applications