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Fthe . Wnen = None 1 V 1 penonning sionicanc cnoices 1 cotnec #peciticd 1 Va ue9s gnincanl...

Question

Fthe . Wnen = None 1 V 1 penonning sionicanc cnoices 1 cotnec #peciticd 1 Va ue9s gnincanl

fthe . Wnen = None 1 V 1 penonning sionicanc cnoices 1 cotnec #peciticd 1 Va ue 9 s gnincanl



Answers

$\mathbf{v}_{1}=(0,1,-4,-1), \mathbf{v}_{2}=(3,5,1,1)$

This question gives us a sequence and asked us to determine a formula. What we know that the numerator for each term is one, and the denominators essentially add the numbers after each other times, too. So, as we said, the numerator is one and the denominator times two minus one and this works. If you plug in 1 to 3, you'll got 1 1/3 1 fifth onwards.

In this video, we're gonna go through the answer to question of the 19 from chapter 9.4. We have to find for which values that t x one x to the next three are linearly independent. So that's be linearly dependent. Then that would be a nontrivial solution for C for Constant C wants to see 32 this equation, which is just a linear combination off the specters as equal to zero for Okay, So the middle component of each of these back zeros that's not gonna give us any mission to talk. That first and third component of each of the vectors is the same. So we can rewrite this vision as C one close T's crabs. That's I see one plus T c two plus t squared. Psi three is equal to zero. Okay, so it's easy to see that this equation, um, can only be satisfied with nontrivial. See, once easy, easy. Agree. If tea is full of what? Because ah, one tea on dhe t squared. So one T and C squared linearly independent. Less t is equal to one. So x one x to the next three. Uh, Millie Independence, uh, t being any value between minus infinity, infinity except value. Well,

In this video, we're gonna go through the answer to question number 19 from chapter 9.4. Where were asked to find Verify whether the vectors x warm next to the next three. Ah, literally independence for linear dependence on thean Faulty between minus infinity. Infinity. Okay, First, let's try two different, um, two options. So first, if t is equal to zero, that is pretty clear that we can, right? Thanks Two. Well, in this case, X two would equal X tree so we can clearly right, but dead linear combination off each of them of x one x to next three equals zero with nontrivial constants. Could just do it here because next to our next Thio equal to each other next up, if t is no equal to zero, then x two is it could just be written T times 101 which is just tea lattes off x. Well, And since tea isn't zero, we've written a linear combination off ex water next to with non trivial Constance. So therefore, uh, for T Nazi heat with zero, then linearly dependent as well. So always linearly dependence

They're. So for this exercise we have this vector B. And the subspace dovey generated by the one, V two and V three that are these vectors that are defined here. So basically we need to calculate the Earth a little projection of you on this space to view. And just remember remember this projection is calculated as the inner proud of the vector V. Each of the generators of this subspace dog. In this case the generators RV one, The two and 3. So we need to calculate the we need to calculate the inner part of me with each of the generator divided the score of the norm of the generators times degenerates. So these for the three vectors B two square plus the interpreter of B would be three. B three. Did the square of the norm of B. Three. Okay, so just to remind you a little bit of the geometric intuition of this, is that the view is generated by these three vectors. So what we're doing is projecting we on each of the generators and then some that together. So we want We t. v. one and V three acts as a basis. Actually in this case they are linearly independent so they form a basis for this. Yeah, subspace of you. So we're writing the in terms of this basis. So we're projecting projecting on this sub space. So let's calculate the correspondent values that we need. So in this case we would be one. The product of B would be to dinner product of the would be three. So this is equal two, one half, There is a constitute and this inner product is equal to zero and then the norms. So because this is the cost to zero means that we don't need this term anymore is going to be equal to zero. So we just need to calculate the score of the norms for B. two and B one. So for me, one square of the norm, remember that there is equal to the inner product of the vector with itself. And in this case this result in one and the inner approach of B two square is equal 2, 1 as well. So these are actually military vectors. And then we just need to put all together on the four. So behalf that the projection of the vector B on the subspace, our view, it's equals to 1/4 times 11 one plus the vector V two. That is equal to one, 1 -1 -1. After some. In these two vectors obtain the action solution that is one half times the vector, three, three minus one minus one. That corresponds to their thermal projection of beyond this subspace of you.


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