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Question

Let $H$ be the set of all vectors of the form $\left[\begin{array}{c}{2 t} \ {0} \ {-t}\end{array}ight] .$ Show that $H$ is a subspace of $\mathbb{R}^{3}$ . (Use the method of Exercise $9 . )$

Viswesh Uppalapati · Accepted 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

Daniel Pezzi · 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.

Donald Albin · Answer

The question says, Let s be the subspace of three dimensional real space consisting of all vectors off the warm V is I see one. See to C to minus two. See one, determine a set of vectors that spans s well, the first vector could be where C one is one NC two a zero zero minus. So I&#x27;m gonna write it in here. Zero minus two times one 10 Negative too. The second vector could be where C two is one and see 10 zero one one minus two Time zero So V to a vector would be 01 one. So these two vectors would span vector space s.

Daniel Pezzi · Answer

Hello For this question, we&#x27;re asked to show, uh that the set of all matrices that could be represented. This form where A, B and D are just arbitrary real numbers is a subspace of em two by two, which is the set of all two by two matrices which in and of itself is a vector space. So we need to check the three subspace criteria. So the first subspace criteria is that the zero matrix, which looks like this is this within H. And the answer is yes, because I can just set a equal to be equal to D equal to zero. And clearly the zero matrix matrix is within H. So I give myself a little check mark here. Secondly, if I add two elements of h do I end up back in h and just a little quick, um, trivial calculation. Remembering the the definition of Matrix edition shows me that, yes, this is in fact, true. And for these types of questions, where you&#x27;re asked, toe justify that something is in fact, a subspace. It should be fairly routine. Um, just checking these three criteria and a lot of times it&#x27;s just writing down things that seem, um, incredibly obvious and because their field is the real numbers. But I had to real numbers. I end up back in the real numbers. Eso this is in H and I get another check mark And then for the third one, If I take an arbitrary constant, I&#x27;ll call it, um Alfa. That&#x27;s a That&#x27;s a real number and I multiply it times and arbitrary member of my set Do I end up with in a church? And of course, the answer is yes, because zero will cancel out, um, any constant that I like. She&#x27;s clearly within H because we were operating with the field of the real numbers. So Alfa Times a Alfa Times being Alfa times D are all real numbers. I get my third check mark, and I&#x27;m done. H is, in fact, a subspace of M two by two

Runpeng Li · Answer

problem. We&#x27;re given age, which is son and dimensional. Subspace, subspace off. Hey, which is also and dimensional. And where will we want to show ages people to be so first to Christine to notice that that stuff space age as the basics with the in vectors and those in vectors has to be the new independent. So already down here. So that is Asia has a basis. Oh, and electors, because age it&#x27;s in dimensional and more over those vectors has to be leaner. Independent, no. Are we nearly independent? Being dependent now spelled it wrong. Independent Dent. Okay, so Oh, I forgot to say that these factors has to be, you know, the independent in B because H e&#x27;s a subspace off. All right, so now we can apply bassist zero, you know, a textbook, these this serum. So it says by our basic faces, the room. So they, the vectors will constitute well constitute a basis for yes. Well, so factors constituted he two, right? So that means issue. Because why? Because Asian each and b r o r both spend by the same set of vectors. So there&#x27;s no other way to find a dis joint Asian V to make this happen. So this is our conclusion. We just have a vectors, that kind that span, uh, both H and visa. These two spaces has to be his two spaces have to be.

Problem

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

Problem 10 Medium Difficulty

Let $H$ be the set of all vectors of the form $\left[\begin{array}{c}{2 t} \\ {0} \\ {-t}\end{array}\right] .$ Show that $H$ is a subspace of $\mathbb{R}^{3}$ . (Use the method of Exercise $9 . )$

Answer

$v=\left[\begin{array}{c}{2} \\ {0} \\ {-1}\end{array}\right]$

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

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Kristen K.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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$v=\left[\begin{array}{c}{2} \\ {0} \\ {-1}\end{array}\right]$

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Kristen K.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Anna Marie V.

Kristen K.

Joseph L.

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

Linear Algebra and Its Applications