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Problem 13: Consider the basis B = {v1 = [1,2,31T = v2 = [1,2, 0JT , vg = [2, 0, 0JT} for R3 . Find the coordinate vector (v)B for the following vectors:v = [1,0, 0...

Question

Problem 13: Consider the basis B = {v1 = [1,2,31T = v2 = [1,2, 0JT , vg = [2, 0, 0JT} for R3 . Find the coordinate vector (v)B for the following vectors:v = [1,0, 0]T 8 U = [0,1, 0JT U = [0, 0, 1]TUse your answers for parts (a)-(c) to find the change of basis matrix AB'-B that sends a vector & (represented using the standard basis B') to (w)B.

Problem 13: Consider the basis B = {v1 = [1,2,31T = v2 = [1,2, 0JT , vg = [2, 0, 0JT} for R3 . Find the coordinate vector (v)B for the following vectors: v = [1,0, 0]T 8 U = [0,1, 0JT U = [0, 0, 1]T Use your answers for parts (a)-(c) to find the change of basis matrix AB'-B that sends a vector & (represented using the standard basis B') to (w)B.



Answers

Find the change-of-basis matrix $P_{B} \leftarrow c$ from the given basis $C$ to the given basis $B$ of the vector space $V.$ $V, B,$ and $C$ from Problem 19.

Get for this problem. They want us to find a basis or a subspace of square major cities by two rial entry. The subspace is all matrices with trays. Zero. What do those look like? Well, that's right Now and B. C Well, if the trace is going to be zero at this point, we have to have negative. Is this an injury? And now let's This is an arbitrary element of our space for any choice of ABC. But now let's write our main Chrissy in a different way. Negative. A 00 You can write as the some of these three mantra sees 000 seeing plus 0 +00 be Okay, Well, now we can throughout the scaler. No, since the one ce time 0001 plus spee times +0001 And so we see the this arbitrary element of us has been uniquely written in terms of these three major C and they do form a basis for our subspace. Okay, Awesome.

Okay for this problem, They give us the vector space of all any year combinations of three functions. F of X is equal to either the ex two of X's equity to the negative X. And if three of X is equal toe sign hyperbolic Scient of X. Okay, so we're going to generate new functions from these in terms of linear combinations, so we can take any choice of real member a times of one of it, plus any toys around number B times after of X plus an in choice of real number seat times. Have three of X and get some new riel function again. But these three aren't linearly linearly independent. Why is that? Well, three of X hyperbolic sign, you may remember from your heavy Rolex I in identities never lose that flash Garrity wrote a long time ago into the X minus e to the negative over to Well, let's see that just 1/2 eight of the act minus 1/2 either the negative eggs. So that's just 1/2 times that one of eggs attract 1/2 times aftereffects, so after three is like the set is linearly dependent. Every can be written as a linear combination. It's in the span of the 1st 2 function were viewed as vectors. Only two of these vectors are a basis are needed to form a basis, and we just choose F line and after two.

The answer to this problem is the embarrass off the metrics off in problem 17. But to solve it separately the fresh right, The first picture off BC's see Because to And then we need to find a better C one C two that been multiplied by the victors off B gives us, uh, first victor off. See, I mean, he every Red Sea nine to close see to for at three. So here this is X. This is why and then see one bye bye except the first si two foot by takes a second and keep this X. So she's not equations here. Nine c one plus for C to give this to on to C one plus series Negative self minus three C to you. Just one. Then we need to solve these. I just defying games. You can write c one musical, too one plus three c two divided by two and then replace see one here and then and gets you want a C two she's going to be write it as tell you netted two basis be to seventh and negative one. Yeah. So this is the first vector I need to find a second and we write the again. Victor off basis. See 83 in one. Go see one 92 plus C two for a negative three. I seem likely. Solve it. Um, like above. You're right. This is the equations. Nine C one plus for C two. Close to negative. Three to see Yvonne minus C three if you to you calls to one, then here. You can't just simplify as before. See? One used to call to one plus three C two you wanted by two on. Then replace. See one here, get C one and C two. The second victor will be as to be minus 1/7 on three minutes. So these two pictures are the columns off. Um, the change of metrics. Change of basis, Metrics. So hairy. Right? The the metrics from here. Noise from C to B. Easy call to the metrics. Seven. Maybe so just be joke metrics from C to B

Yes. For this question, they I want us to find the basis of the pond on the house of degree here less when all of the elements in that basis let's call that speak, have the same degree. So I'm gonna make them all have degree Three, actually. Only need to do this. They can't all have agreed to will never be able to get the cube term. But I'll make it simple still, uh, this one x cubed plus x squared x cubed plus X. Why is that? Last time I didn't even have to bother translating them. Teoh, Um, a system of equations and our fort after translating their coefficients to appropriate vector injuries. You can do that on your own if you want, but and actually you're gonna correspondence between the algorithms on our for and these basis vectors right here. So they want us to prove their independent. Remember, we would do Gaussian elimination when trying to solve for the zero vector. But here we can keep our language in terms of polynomial. There's going to need to be some coefficient times this that ends of Equalling zero again. What we're doing, we're taking these vectors on resolving for the zero factor bombs. Girl Too much. Let's see. Scroll! Thank you. And then we have and three Let me write these coalition Let's actually do plus x Yeah, Ah, one a three and then our second basis selecting Let's see if this is also anyone. Yeah, A to X cubed plus a two x squared. Okay, So scaler multiplied by this affecting a different scaling a three multiplied by this sector over here plus one times a three and then lastly it one more Let me space out zeros in a four times x cubed. Is that what says your well, if he translated this into a system of the nuns a 1234 I'm interesting. It's kind of almost and go see and form already. So it's really nice to translate the solutions. The only solutions A for that makes the zero is a four is equal to zero. Let's see more level and the same thing here. The only solutions of this would be if Avery was zero. You can factor out a three year contract out they do you hear the A one here? They all have to be zero and we proven in this manner that they're independent so they aren't basis


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