could-a-set-of-three-vectors-in-mathbbr4-span-all-of-mathbbr4-explain-what-about-n-vectors-in-mathbb

Name: could a set of three vectors in mathbbr4 span all of mathbbr4 explain what about n vectors in mathbb
Uploaded: 2019-08-21
Duration: 2 min 24 s
Channel: Numerade
Description: Could a set of three vectors in $\mathbb{R}^{4}$ span all of $\mathbb{R}^{4} ?$ Explain. What about $n$ vectors in $\mathbb{R}^{m}$ when $n$ is less than $m ?$

Question

Could a set of three vectors in $\mathbb{R}^{4}$ span all of $\mathbb{R}^{4} ?$ Explain. What about $n$ vectors in $\mathbb{R}^{m}$ when $n$ is less than $m ?$

Runpeng Li · Accepted Answer

Okay, so for a problem 32. Um um, we&#x27;re asking coulda set off three vectors ing our force been all our four. So the answer. Excuse me? The answer is definitely no or, um yeah, that the answer is definitely no. The reason is that if we want, we want to want want some some base factors to spend. They are for the least number of factors we need is four vectors. So you can just think about, uh, such a vector like, um 1000 100 And I guess that I think about such a such a basis, like 0100 and 0010 and 0001 So these four vectors spend our four. But if we delete any one of them, then the best that we can get is the are three women are three space because we lose a will, use a vector that and will define a one more. One more dimension. So to find out the span off of her four dimensional space, the least vectors we need is four vectors. So three directors will not be enough. So what about, uh what about in vectors for all right, our m within smaller than m again This again Not It&#x27;s not possible. The reason is the same. We need at least I mean at least am vectors to spend the whole space off our r m. Because with the dimension of em, we we have to we have to make m vectors to divine am dimensions. So if we only have in vectors, then we cannot divide every single dimension off our M. So this is also not possible.

Donald Albin · Answer

the question, states. Let s be the subspace of three by three matrices with riel coefficients consisting of all three by three symmetric matrices Find a set of vectors that span eso the symmetric matrices could look like this ABC see No b a de See De Yeah, that&#x27;s good back. It doesn&#x27;t have to have in a there. ABC de you that has to be. See, that has to be e f. Okay. There. Yeah. Um all right, so that&#x27;s the general form. So a set of vectors that spans s we can just write it as first. We&#x27;re gonna have a B one. Everything else his zero. Then we&#x27;re gonna let be B one. Then we&#x27;re going to let CB one. Then we will let db one. Then we will let e b one and everything else zero. And finally we will let f be one. So there you have it. Six matrices that all in and of themselves are symmetric and span the subspace

Donald Albin · Answer

the question reads. Recall that three vectors v one v two Envy three in riel, three dimensional space are co plainer if and only if the determinant of the of a matrix of the three vectors is zero. Use this result to determine whether the given set of vectors spans three dimensional real space. So one common negative one comma one. These were the three of actors two comma, five comma three four common negative two comma one. It says on page 276 in your textbook that any three non zero and non complainer vectors V one, V two and V three in real three dimensional space will spend riel three dimensional space. So I hope I read that correctly Any three non co plainer of actors. And so, um, it we can show that these three vectors are not co plainer. Then they do span three dimensional space. So the determinant of V one v two v three um, Vector one is one negative 11 vector to is to five three. Vector three is four negative too one. And I&#x27;m going to figure the determinant by writing the 1st 2 columns out again. Okay. And then one times five times one is five negative. One times three times four His negative. 12 Naor. One times, two times negative two is negative four. And now I have to put minuses in front of the next ones. One times five times four is 20 one times three times negative. Two is negative. Six minus negative. Six is plus six negative. One times, two times one is negative to the opposite of negative two is positive. Two we add all this together. Um, I&#x27;m just gonna put this in a calculator. Five minus 12 minus four minus 20 six plus two gives us negative 23. So since the determinant is not zero, then these three vectors are not complainer. They are non complainer. So the vectors span three dimensional real space.

Donald Albin · Answer

This is the same question as, um problems five and six. So we have three vectors. Two comma, negative, one comma, four, three common negative. Three comma five, one comma, one comma three. And basically we need to figure out whether these vectors are co plainer. So if we take the determinant, then we will know if the determinant is zero than the vectors are co plainer. I like to calculate the determine it by writing out the 1st 2 columns for this graphical method. Gonna make sure I didn&#x27;t write anything incorrectly. Always a good idea to double check your work. Except maybe on a test when you&#x27;re really pressed for time. All right. Looks good. Two times negative. Three times three. Um, it&#x27;s negative. 18 negative. One times five is negative. Five. Four times three is 12. I need to put a minus in front of the rest of these negative four times. Three is negative. 12 negative negative. Three times for his negative. 12. So minus negative. 12 is plus 12. Two times five is 10 negative. One times, three times three. That&#x27;s going to give me negative nine minus negative. Nine is gonna be plus nine Okay, I&#x27;m gonna put this in a calculator. Negative. 18 minus five plus 12 plus 12 minus 10 plus nine zero. Because the determinant zero, then these vectors are co plainer. And so, um no does not span rial three dimensional space.

Donald Albin · Answer

give a geometric description of a subspace of riel three dimensional space spend by the given set of actors. So the given set of actors is just V one. So that&#x27;s only one vector where V one is a non zero vector. So that&#x27;s just one vector. And so a description would be, um, whatever that direction that vector is going, it could be lengthened. Or you could multiply it by a negative and you could make it Ah, opposite or lengthened. And so it would be the points along a line in three dimensional space, following the direction of Vector V one.

Daniel Pezzi · Answer

Hello. For this question, we need to find the dimension of a space which I&#x27;ve just arbitrarily called W, which is the set of all vectors in our three where the 1st and 3rd entry are equal. So if we look at this this Vek, this vector that I&#x27;ve written a b a all vectors which are in our space w could be written in this form. We can break it up into a times a vector plus b times a vector. On those vectors are 101 and 010 Which means that, uh, W is actually all of this implies that w is the span of two vectors 101 and 010 And this immediately implies that the dimension of W is to because this span of any set of vectors is a vector space whose dimension is equal to the number of linearly independent vectors and span. And clearly these two vectors air literally independent eso We know that the dimension of W is to another way to think about it in a less precise sense is that if the 1st and 3rd entries are equal, then I have two choices for my Constance because I can set the second slot, whatever I want. And I can set the first slot, whatever I want. But that forces the third slot to be, ah, specific value. So the dimension, the number of choices I have for Constance is said it too. But this is, ah, more formal way of expressing that idea. And it also gives you the two vectors, um, that represent this factor space and that.

Problem

Suppose $A$ is a $4 \times 3$ matrix and $\mathbf…

Problem 32 Hard Difficulty

Could a set of three vectors in $\mathbb{R}^{4}$ span all of $\mathbb{R}^{4} ?$ Explain. What about $n$ vectors in $\mathbb{R}^{m}$ when $n$ is less than $m ?$

Answer

Same thing for $n$ vectors in $\mathbb{R}^{m}$ . Than matrix can have at most $n$ pivot
$\left.\text { columns, so it can span } \mathbb{R}^{n} \text { but not } \mathbb{R}^{m} \text { (because } n<m\right)$ .

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

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Heather Z.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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Same thing for $n$ vectors in $\mathbb{R}^{m}$ . Than matrix can have at most $n$ pivot$\left.\text { columns, so it can span } \mathbb{R}^{n} \text { but not } \mathbb{R}^{m} \text { (because } n<m\right)$ .

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Heather Z.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Anna Marie V.

Heather Z.

Michael J.

Same thing for $n$ vectors in $\mathbb{R}^{m}$ . Than matrix can have at most $n$ pivot
$\left.\text { columns, so it can span } \mathbb{R}^{n} \text { but not } \mathbb{R}^{m} \text { (because } n<m\right)$ .

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

Linear Algebra and Its Applications