In exercises 1 6 the linear transformations s t and h are...

In Exercises 1-6, the linear transformations S, T, and H are defined as follows: $S:P_3 \rightarrow P_4$ is defined by $S(p) = p'(0)$. $T:P_3 \rightarrow P_4$ is defined by $T(p) = (x + 2)p(x)$. $H:P_4 \rightarrow P_3$ is defined by $H(p) = p'(x) + p(0)$. 1. Give the formula for $S + T$. Calculate $(S + T)(x)$ and $(S + T)(x^2)$. 2. Give the formula for $2T$. Calculate $(2T)(x)$. 3. Give the formula for $H \circ T$. What is the domain for $H \circ T$? Calculate $(H \circ T)(x)$.

Transcript

00:04 Hello there.

00:06 Okay so for this exercise we have two transformations one from the space of polynomials of at most degree one to this piece of polynomials of at most degree two and then we're going to pass from p2 to p3.

00:19 Here is defined the corresponding basis for each of these spaces b b prime and b double prime the transformations are defined as follows we in the case of the first transformation here we are taking a polynomial and then we return of this form c0 plus c1 x and we return to z0 minus 3 c1x.

00:44 And the second transformation, what it does technically is just multiply by 3 the polynomial and multiply by x as well.

00:54 Okay, so we need to find the corresponding matrices for these transformations relative to the corresponding basis.

01:07 So first we need to start with this basis b and b double prime for the first transformation.

01:14 So let's begin with the first transformation and let's find this matrix.

01:22 Here the idea is that we need to pick the basis from b and transform it using the rule, the transformation t1.

01:31 So t1 here is the fine big polynomial z0 x1 x and it returns 2 times z 2 z0 minus 3 times z 1 x okay so then we need to apply this transformation to each of the element basis first to 1 so we applied t 1 to 1 and we obtain in this case and for t2 apply to x 41 applied to x well it returns minus 3x now the idea is that we need to represent this on the on the basis b b prime that is this basis here because this is transformed to b prime basis and then we can represent this as a vector to zero zero and the same for the second one this will correspond to the to the vector zero minus three zero of course this is in the basis b double prime b double yeah here typo here is this b prime and this b double prime okay so that's represent the first in matrix and then we need to find the second one.

03:22 So for the second one we need to do the same.

03:27 We're going to map now from b prime, double prime to the basis b prime.

03:38 So now we need to consider b a double prime.

03:40 This basis is the standard basis for b2 and the transformation in this case, t2, is c2, is c2.

03:52 0 plus c1 x plus c2 x square and this equals to 3 z0 x plus 3 z1 x square and plus 3 z2 x cube here then we need to apply to each of the elements in the basis so t2 apply to 1 it's just 3 t 2 apply to 1 it's just 3 2x is equal to 3x square this is x 3x and finally t 2 of x squared of x squared is equal to 3x cube so you you need to represent this on the basis in this case b prime so that means that this vector here this 3x this polynomial this monomial is the vector 0300 in the basis b prime.

05:02 For this monomial we have something similar in this case is 0 030 b prime in the basis b prime and the same for the last monomial here.

05:16 This is just 0 0 0 0 .03 in b prime...

Question

Please give Ace some feedback