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Question 5Let v1, 02, U3 and w1, W2, W3 be vectors in V. Suppose (v1, 02, U3) spans the whole space V, and (w1; w2, w3) is linearly independent:Is the following sta...

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Question 5Let v1, 02, U3 and w1, W2, W3 be vectors in V. Suppose (v1, 02, U3) spans the whole space V, and (w1; w2, w3) is linearly independent:Is the following statement true?(v1, 02, U3) is linearly independent:Your True False selection(Only for true false question)true

Question 5 Let v1, 02, U3 and w1, W2, W3 be vectors in V. Suppose (v1, 02, U3) spans the whole space V, and (w1; w2, w3) is linearly independent: Is the following statement true? (v1, 02, U3) is linearly independent: Your True False selection(Only for true false question) true



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In Exercises 29 and $30, V$ is a nonzero finite-dimensional vector space, and the vectors listed belong to $V$ . Mark each statement True or False. Justify each answer. (These questions are more difficult than those in Exercises 19 and $20 .$ )
a. If there exists a linearly dependent set $\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\}$ in $V$
then $\operatorname{dim} V \leq p$
b. If every set of $p$ elements in $V$ fails to span $V,$ then $\operatorname{dim} V>p$ .
c. If $p \geq 2$ and $\operatorname{dim} V=p$ , then every set of $p-1$ nonzero vectors is linearly independent.

In this problem, we need to prove that if anyone V two is linearly independent set of vectors and the three does not lie in the span of you on V two, then we one, V two, V three is a set of linearly independent vectors. Now, for this, first of all, note that since he wants me to is a linear independent set of vectors, so we one cannot be written as a scalar product of V two. So that means that everyone is not equal to K times V two for any real number key. Also, it is said that the three does not belong to the span of We want V two. So this means that we tree cannot be written as a linear combination of V one and V two. Now, next suppose that it is possible to write V one as a linear combination of V two and V three. So we should be able to write V one is equal to K two, V two plus K three, V three. Where K two and K three are real numbers. Is that both of them are not equal to zero? Now, if we assume that the three is equal to zero, then in that case we want will be equal to key to be too. And this will contradict the fact that everyone cannot be written as a scalar multiple off V two. So this means that K three cannot be equals to zero and hence K three must be not equals to zero. Now, since K three is not equal to zero, we can rewrite this equation as key three. V three is equal to be one minus K two, V two. Which implies that we three is equal one divided by key three times we want minus key to divided by key three of V two. We can do this since K three is non zero. But here we can see that the three has been written as a linear combination of V one and V two. And this will contradict the fact that the three does not belong to the span of we want to. So this means that the one cannot be written as a linear combination of V two and V three. This assumption is wrong. So in a similar manner, we can prove that we too will not be able to be written as a linear combination of the one and V three. So what we obtain is that the one cannot be written as a linear combination of V two and V three. We do cannot be written as a linear combination of V one and V three and V three cannot be written as a linear combination of V one and V two. We get this from the first part. Over here. V three cannot be written as a linear combination of V one and V two. Later on, we prove that we won cannot be written as a combination of V two and V three. And we also show that we too, cannot be written as a linear combination of V one and V three. From all of these. We can conclude that we want, we do V three is linearly independent set of records. Yeah.

Yes, So here. So we want to argue. That's so B one b two B three before it's Kenya independence. So first we consider the linear combination Nike. So let's do a one B one plus a two b two purse. His dream. This we just a four before it's the country barrel and we consider. So if they're leaning into pensions, the wisher conclude that all the, uh or the A's should be zero. So the first thing is, if a four, it's Nico Todaro. So if If so, our for Miller reduced to released director right into according to the condition we already know B one B two b sweetie is leaning independents, so we can deduce a one. We go to a to be continued Ansari. It's Kiko Trujillo. So so here, every since zero for the second. So if a four is not Yuko Todaro, so then where happened? So then what happened is if we can write a four before it's minus a one B one us into B two is the three is because it's not there. We can do by the the right hand side by a four. So before its miners, he went over a four. Be one minus into over a four meets U minus a three over a four b. C. So here we can see before kids that spin by, we want to research. That's the condition. Tear us before it's nothing. The span. So this means this condition is not true. So a four had to be zero. So if a four had to be there Oh, all the coefficient had to be there. That this means if we have the linear commission and also coefficient had to be Barrow, then we know B one b two b three before I leave me needing chips.

Here. This state's proved that if the set containing vector V one in Vector V two is linearly independent and if V three is not in the span of the vector of you one in Vector me too. Then the set of vectors, v one, V two and V three is linearly independent. So if it's not in the span, then Vector V three does not equal any constant times. Vector V one plus any constant times Vector V two. All right, so Vector the three is not a linear combination of vectors V one and vectors v two. And so we could right that if we want the set of vector V one, Vector V two and Vector V three to be linearly independent. That means that some constant has vector view one plus some constant groups times vectored v two plus some constant times. Vector V three cannot equal zero. All right, so I'm gonna try to prove by contradiction. So I'm going to let d one v one plus d two v two And you know what? I better write this over again. That t one v one Plus it d two v two plus de three V three equal zero for some really numbers, do you one de to Andy three. If that were true, then by subtract d three v three from both sides of the equation I get and then I divide by negative d three on both sides. I would get the three equals negative D one over de theory V one, minus D two over d three v two. And if I, uh, say that d one over D three is some other constant C one and I mean negative, do you? It went over to three. And if I let negative de through to over 83 b some other constant C two then V three is C one V one plus C two v two. But we already said that V three does not equal. See one V one plus C two V two for some C one C two that are really numbers, which is a contradiction. And so that means therefore d one v one plus de two v two plus de three v three do not equal zero the opposite of our initial statement

All right, We're gonna be doing problem 29 in section 4.5 at the beginning of problem 29. It says ve is a non zero. So we have a vector space that's non zero and finite dimensional. All right, So if V is ah non zero finite dimensional vector space and the vectors listed below belong to be, then the question gives you three statements. Asked you whether or not they're true or false and to justify those answers. So we're gonna start by considering a If there exists a set V one to V. P. So let's write down the set that we have V one to V. P that spans V. So this is ah, spanning set. So the sets fans be, then the dimension a V is less than or equal to P. Now we're going to solve this by considering two different types of sets that could span V. First, we're gonna consider a linearly independent spanning set. So if you have a linearly independent, spanning set that is also known as a basis and if it is a basis than the dimension V equals P, this is because none of your vectors from V one to V. P could be written as a linear combination of any of the other vectors. Therefore, that would make dimension of be equal to P. Now let's consider the other option of a linearly dependent set A linear, linearly dependent, spanning set would not be a basis and therefore the dimension of the would be less than P. This is because we could have more vectors in our set than would be needed to form the basis of the Therefore, the dimension of the would be smaller than the number of factors we have in the set. So therefore this one is true and there's your justification. Give me just a moment to erase all of this. We will start on beef. Alright now for be that says if there exists a linearly independent set V one to V p. So we have you want to VP and this set is linearly independent, linearly independent. Then your dimension of the greater than or equal to P. Now we're going to solve this one or justify this one in basically the same way we did is a We're gonna consider two types of linearly independent sets. We're going to consider it. This linearly independent set spans V So if it's a spanning set, as we said before, a spanning linearly independent set forms a basis and if it forms the basis, the dimension of the equals. P, however, say that this is not a spanning set. This is not a spanning set that's going to make the dimension of the be greater than P. This is because there will be fewer vectors in the set V one to V. P. Then there would be needed to form a basis. Since the set does not span. Therefore, the dimension of the it will be greater than the number of vectors in P. Therefore, he is also true. Give me one second to a racing will go on to see alright now for see it gives you. If the dimension of the equals p there should be a spanning, set spanning set of P plus one vectors envy. We already know about the existence of linearly dependent, spanning sets, which we already know would make this true because of linearly dependence spanning sets. But let's just give a quick example for this one. Let's say V equals our sweat the dimension of e equals the dimension of r squared, which equals two. This too, is equal to our key. And now let's give a spanning set that has P plus one factors in it. No, this is a spanning set and there exists P plus one vectors. Therefore, this one


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