in-exercises-15-18-find-a-basis-for-the-space-spanned-by-the-given-vectors-mathbfv_1-ldots-mathbfv-2

Question

In Exercises $15-18,$ find a basis for the space spanned by the given vectors, $\mathbf{v}_{1}, \ldots, \mathbf{v}_{5}$
$$
\left[\begin{array}{l}{1} \ {0} \ {0} \ {1}\end{array}ight],\left[\begin{array}{r}{-2} \ {1} \ {-1} \ {1}\end{array}ight],\left[\begin{array}{r}{6} \ {-1} \ {2} \ {-1}\end{array}ight],\left[\begin{array}{r}{5} \ {-3} \ {3} \ {-4}\end{array}ight],\left[\begin{array}{r}{0} \ {3} \ {-1} \ {1}\end{array}ight]
$$

Doruk Isik · Accepted Answer

and this problem, we&#x27;re given a sending electors knew where to find the basis for this that so let&#x27;s form our system for big from the given lectures. Endless side of this matrix site is our system a CZ you can see hey has four rows and by colors. So it means that a mite hair enough given elements for Pete Rose, since the number of problems is greater than the number of rows. So we know that this system will spend the space. Um, next thing to do is to find the reduced role for or by this system and that Is he going to 1000 0100 000 Sorry. 0010 Negative one negative. 300 Negative to link two by two and zero. Since we know that this settled vectors, I&#x27;ll spend the space Don&#x27;t be basis. Dundee column Give its columns will be the basis for the system. Hey, what are you giving calls? The 1st 1 the 2nd 1 and the 3rd 1 will be our give. It holds so many faces where we set a will be the first told him off a 1001 The second column Obey negative to one negative one. And her column? Um A, which is six negative one, too.

Michael Jacobsen · Answer

in this example, we have five vectors provided and what we&#x27;d like to do is define a vector space h to be the span of vectors, v one, V two, V three before and V five. So now the H is the span of this set of vectors. One thing to notice is that the set here does not form a basis. The reason being is that we have five vectors, but only four entries per vector. And whenever we have more vectors than entries per vector, the set here would be linearly dependent. So we know it spans. And if we want a basis, what we need to do is to decide which vectors to delete so that we still span h and form a basis. And as you might have suspected, the trick to determining all of those things at once is to form a maze tricks and ro reduce. So let&#x27;s start off by saying, Let a matrix a be given as the one be to the three before and be five as its columns. Then this matrix is going to be 10 negative. 32 012 Negative three Negative three negative 416 and one negative. Three Negative 87 last. It&#x27;s 21 negative, six and nine. So this is going to be possibly an interesting role reduction. Since we&#x27;re dealing with a four by five matrix, let&#x27;s see what we get into. We&#x27;re only going for echelon form, let&#x27;s remind ourselves. And so from this pivot, there&#x27;s just two inch, two issues We need to eliminate this and this entry. So here I&#x27;ve copied the first rows of this matrix and now let&#x27;s multiply row one by positive three At the results to growth Row three, what will obtain is a zero to negative eight negative five and zero for the next step, multiply row one By this time a negative two and then add the results to the last row. This operation will give us all together a zero, then negative three Looks like a positive 12. Then a five and a five. So with analyzed, this new matrix we have a pivot. Here was euros below. We have a pivot here as a zero above, but keep in mind, we&#x27;re just going for echelon. So that&#x27;s nice but not needed. But we need to eliminate the two and the negative three. So let&#x27;s start by re copying this matrix Rose ones and two. Mm. So the matrix is copied and we know have a zero here and here. So let&#x27;s look at what the pivots are. This time we have a pivot and column one a pivot and calm to as well as a pivot in Column four. And that&#x27;s all we have. But I also noticed that column one correspondent to V one. Then we have the two and V four. So what we need are V one V two, Envy four If we collect these particular vectors where you will still span, but they will also be linearly independent, since they corresponded to the pivot columns and a reduced echelon form. So we&#x27;re ready to state our conclusion. The set consisting of V one V two and before we already know it spans because that was given. But we also know it is linearly independent now, by definition, a set of vectors that spans and is linearly independent forms a basis. So we can say, therefore, this set forms a basis. We should say a little bit more, and we have to be careful. It doesn&#x27;t form the basis for our four. It forms a basis for what its fans, which is H. It&#x27;s just a subset of our four, so be sure to say it forms a basis for H itself.

Michael Jacobsen · Answer

in this example were provided with five different vectors and each of them come from the set are five. What we&#x27;re going to do with this set of vectors is let H be equal to the span of B one B two, b three, V four and V five. So now h is a subspace of the larger space are five. Merkel here is to form a basis for this speech. The space age in forming a basis. We have two requirements. One requirement being that we spend the space we require. And so that&#x27;s already met. We already know the one through the five span H because that&#x27;s how we defined age. So now we&#x27;re looking for a set of vectors from this which are linearly independent. They may all be linearly independent or just a collection of them. And to determine this, we&#x27;re going to use the following method. Let a matrix a be formed from columns being the one B two, b three, V four and V five. Now our method is to row, reduce and find which columns are pivot columns, but unfortunately reducing a five by five matrix with not the most friendly numbers like we have is a tremendous amount of effort. So what we&#x27;re going to be using is a calculator to do the job for us. So that way we can save some time and get back to linear algebra ideas. So in a calculator, I found that this matrix a ro reduces to So now with this matrix, what we need to do our analyze the pivots. But first, let&#x27;s recall that column. One came from V one, then B two, b three, B four and V five. Also, we have a pivot and comes 12 and three, but nowhere else. This tells us that if we want a basis for H V 13 V three will span and be linearly independent. So that tells us the vectors before and the five can be effectively discarded since V one v two V three generate those two as well. So it&#x27;s day our conclusion. We know that the set consisting V one v two v three, is linearly independent. Now recall our Cole. Here&#x27;s to find a basis for the Space H, which is a subset of our five, and we already knew it spans. We found a set that&#x27;s linearly independent, and now we&#x27;re ready to stay a conclusion. This implies that the set the one V to V three now forms a basis. Four. H recall. It&#x27;s not basis for our five as not basis from our three might think with three vectors. It&#x27;s a basis for H because that&#x27;s the set that it spans.

Doruk Isik · Answer

in this problem. We&#x27;re given a set of bikers and no work but as to find the basis or that Viper. So let&#x27;s first for our system that is formed by our vectors and blessed side. This system, as I mentioned size for by pictures, be more B two B three before and fire. So first thing, too. Look at this. We have five rolls, one to triple five. So mentioned saying his five rows and five holes. So since number of rose is equal number of problems, you know that this electors will spend the space. Why is this information important? Well, because since we know that the specter will spend in space, we consider that the given problems off the system will be the basis Becker&#x27;s. Okay, so we&#x27;re actually looking for the basis factors off a So let&#x27;s look at the reduced role equipment for off the system. So then you&#x27;re ready. Tropical form. Our system will look like 10 00001 000 I heard 2000 00100 and or 3rd 1/3 Negative. Once you&#x27;re here, so we help you ever tell him you won&#x27;t have a element here. We have our second favorite element here in the particular element here, so our first you would tell them, would be the first color. Second will be the second column, and our first poll will be the fourth column in the orginal system. A. So then, the basis for this set of electors will be the first column all metrics. A negative 876567 in the second column. Age negative seven Negative mind negative 57 and then the or column, which is 1496 names except

James Chok · Answer

Okay. In this question, we will find the basis of the subspace span by these vectors. So this is a quiver tow asking what is a column space off this matrix when you just squish all of these colored pictures together? So this column back the girls yet some guys here goes here, goes here, goes here. Does it really matter what water it is? But but I&#x27;ll just take a funeral. So then the first thing that we have to do is to find to convert this into right echelon for so the first couple off steps that will do is take a road to you, add one that she wrote to will take Row three and add two tons of road one. In that interview, I three three and then we&#x27;re full. You was tracked. Why? It has wrote one. Bring that into road for so performing the steps 120 minus 131 plus one gives you zero to my street. Get it minus 10 clothes to give to four months. One gives you three, three months. Eight gives you negative five now very three plus words. You so negative Two plus two gives you zero four minus one gives you 30 My sixties Negative six negative seven miles to give me make you mine and nine plus six kids. A kitty. Okay, now rough Full minus 51 So five minus 50 six minus 10. Giving negative pole eight minus zero gives you eight. Seven plus five. Getting too. No. Sorry. Well, and by minus 15 gives you maybe 10. So now, right oppressed That perform is row three and add three. Title wrote to going down into row three and wrote full will subtract four tunnel road to connect to work for So we get 1 to 0 minus 130 Advice 123 Negative five. And you&#x27;re so right. Three plus road to sew. Three plus three miles st zero to negative six plus six is your negative mind plus nine zero 15 minus 15. Now I wrote for my forward to Is there a negative? Four plus four Q zero eight minus eight is you 0 12 miles. Talk gives you zero, but this is a regular for my sake. Zero on dhe negative. 10 plus 20 gives you 10. So you see, we only have three columns, right? Yeah. So how How homes waste is the space to span by These three letters stood in the basis for this subspace. Ah, one minus one minus 25 23 months. 16 and three. Negative, 89 and five.

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.

Problem

In Exercises $15-18,$ find a basis for the space …

Problem 16 Medium Difficulty

Answer

$\left\{\left[\begin{array}{c}{1} \\ {0} \\ {0} \\ {1}\end{array}\right],\left[\begin{array}{c}{-2} \\ {1} \\ {-1} \\ {1}\end{array}\right],\left[\begin{array}{c}{6} \\ {-1} \\ {2} \\ {-1}\end{array}\right]\right\}$

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$\left\{\left[\begin{array}{c}{1} \\ {0} \\ {0} \\ {1}\end{array}\right],\left[\begin{array}{c}{-2} \\ {1} \\ {-1} \\ {1}\end{array}\right],\left[\begin{array}{c}{6} \\ {-1} \\ {2} \\ {-1}\end{array}\right]\right\}$

Linear Algebra and Its Applications

Lily A.

Catherine R.

Caleb E.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Catherine R.

Caleb E.

Samuel H.

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

Linear Algebra and Its Applications