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Question

Find the dimension of the subspace $H$ of $\mathbb{R}^{2}$ spanned by $\left[\begin{array}{r}{2} \ {-5}\end{array}ight],\left[\begin{array}{c}{-4} \ {10}\end{array}ight],\left[\begin{array}{r}{-3} \ {6}\end{array}ight]$

Runpeng Li · Accepted Answer

So in this problem work even a subspace that is spent by these two vectors. Two. Thank you. Bye. And negative for 10 and active. 36 Okay, So in order to find out the dimension, we have to first check whether it is to These three vectors are in urgent ended so we can observe dead. Back to be one. I&#x27;ll just mark B one b two B three here. So the vector bu wan can be can be represented as connective two times of you two because elective two times before. Oh, sorry. Uh, are we read it? It&#x27;s ah neck, too. Two times of you want is for you, too, because an active two times two will be naked for and then to a time Sector five a 10. So, uh, that means view, too. Is multiple offer be one. So the cleaner, the independent vectors are you too? And to be three. So this is our basis. And so that implies our dimension in this case will be too, because there are only two new depend

Jimmy Yao · Answer

Yeah, The permanent one has to gets intermission for this stop space for their P two. So if you could to spin two plus x square into full minutes two X person three x career into one plastics. So when we tested Dimension first, we&#x27;re going to say whether there any inter printed So we have our for two person x square first better for minus two X plus X squared. Yes, calmer. Well, im chest X, you go to their Well, that&#x27;s it, You country there. So here we know. So this is the crib in as I just three big screen my news two bit out scammer mix because to offer a plus comma, you could do the row. So we have The genius system is three because I propose to ST Peter cross the road because you there oh, into to better thus gamma you could to their own into to offer because plus comma, you go to the rope. So if we so that this leniency is turned, we can kids soon. How do we sell this? So like since we don&#x27;t know Yeah, for example one to see. Remember to time that 1st 1 in the minors, sir. One You kids. Six bitter miners comma. You could 20 So then you have this for Miller. So here you can get all the offer. You go to Darrow Better you go to zero in a calm and you go to their up. Right? So this means this this in this the linear independence. So since there linearly independent So that space they spent is have the dimension, see?

Viswesh Uppalapati · Answer

in this video only solving problem. Number 36 of section 4.1 Grand word. Uh, we&#x27;re asked to determine if why is in the subspace of our four spend by the columns of a where why is, um, why is already in our for a zit has four entries given here, and but we need to see if it&#x27;s spend by these specific columns in our four. So in a better part of the matrix, a first we need to set up in augmented matrix here, um, to solve this system. So we just combine the two matrices are to, uh, the Matrix and the vector. So here we have 38 negative five to as the first column. Negative five seven. Negative eight. And negative, too, was the second. Call them negative. Nine Negative 63 Negative nine. Is that their call? And then a line in between. Negative four. Negative eight six and negative five. As the last column as this is the B column in the Matrix equation, X equals B in resolving for a set of scale er&#x27;s X, where any where x one multiplied by the first column and x two multiplied by the second call in the next three multiplied. But third call them would equal why so this season are you have calculator and to make your life easier sense Doing this by hand would take a long time. But, um, so you get if you saw this, you should get and put it in our area. We&#x27;re reduced to ash lawn for me. Should get negative one on DDE 3/5 on zero. So, um, we solved for three values your negative 1 52 53 5th where? Negative 1/5 times. What&#x27;s called this? The one or a one 81 83. So the night of 1/5 time&#x27;s a one plus negative to fifth times a two and was 3 50 times a three equals. Why? So this, um since there is a solution to this system, you know that we know that, um, Why is spanned by these columns the columns of this matrix, and are for

Runpeng Li · Answer

here we have the four Rector&#x27;s. It&#x27;s been such a subspace. That is, 12 zero. Think of 34 one and noted eight, six, five and negatives 30 seven. Excuse me. So, in order to check whether this is four vectors, are you nearly end end? One approach I would recommend is to apply the gushing illumination and check the reduce station warm to see Hi. If there is an theories and then travel and trivial solution. If there is, then that means the&#x27;s four by four vectors are leaner depended. So we&#x27;re done. So Okay, so to do the cash indignation. Oh, just right down First on the first page here and now we&#x27;ll we&#x27;ll keep going to the second page. Okay, so we first views. Second row, plus two, two times. The first road myself. The first rule. So that is fun. Next three, activate connective three, 23 And so the second rule will be zero. But first century and two Sorry. Connected to the second entry. And negative 10 with 1/3 entry, connective six. Or, uh, third entry for the fourth entry. And we&#x27;ll keep the the last roll. Okay, Next, we will try to reduce the the last roll here. We usedto third row plus twice out the third row. Plus, um, that&#x27;s the second rule. Okay, so first row were not changed. So that is one negative three reactivate a negative three. And so second rule was still being active to connective 10 and neck of six. Well, that&#x27;s certainly the third row. We have zero first and cereal with 1/3 entry and two times 7 14 minus six iss eight. All right, so we don&#x27;t need we don&#x27;t even we don&#x27;t even need to redo. Keep reducing, because right now we can We can know whether there are where there is a nontrivial solution. So we first have um x one minus three, x two plus eight Sorry, minus eight. Minus eight x three finest three x four minus three x for is zero. And we have negative too. Thanks to minus 10. Next three finest six x four. It&#x27;s zero and eight times. Explore is always zero. So export zero. But we still have X y extracts three. Right. So first thing we have next one minus three x d&#x27;oh. Minus aid x three. It&#x27;s zero and connected to x two minus 10 X tree. It&#x27;s Darryl well, ex for zero so we can find from thes you questions that x two x three and, um, yaks. Three is a free, bearable, and X two can be determined by X three and X one can be determined by X two and x three. So that means there&#x27;s there exists athletes and nontrivial solution. So that means these four vectors Do you want to three Before are we nearly dependent? Independent dependent. So that implies dimension off the subspace it&#x27;s for.

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

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

In Exercises 11 and $12,$ find the dimension of t…

Problem 10 Easy Difficulty

Find the dimension of the subspace $H$ of $\mathbb{R}^{2}$ spanned by $\left[\begin{array}{r}{2} \\ {-5}\end{array}\right],\left[\begin{array}{c}{-4} \\ {10}\end{array}\right],\left[\begin{array}{r}{-3} \\ {6}\end{array}\right]$

Answer

dimension of the subspace H is 2

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

Linear Algebra and Its Applications

Heather Z.

Kristen K.

Samuel H.

Michael J.

Vectors Intro

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dimension of the subspace H is 2

Linear Algebra and Its Applications

Heather Z.

Kristen K.

Samuel H.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Heather Z.

Kristen K.

Samuel H.

Michael J.

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

Linear Algebra and Its Applications