3 assume p1 p2 p3 p4 p5 are polynomials in the vector space...

(3) Assume p1, p2, p3, p4, p5 are polynomials in the vector space P4(R) such that each satisfies pj(2) = 0, j = 1, 2, 3, 4, 5. Prove that the set {p1, p2, p3, p4, p5} is linearly dependent. (4) Give an example of a linear transformation T: R^4 -> R^4 such that R(T) = N(T). Then show that there does not exist a linear transformation T: R^5 -> R^5 such that R(T) = N(T). (5) Consider the linear transformation from S: P2(R) -> R^3 given by S(f) = (f'(0), 2f(1), 0) in R^3 for every f in P2(R) where f' denotes the derivative of f. Find N(S) and R(S). Find [S]_(β)^(γ) where β and γ are the standard ordered bases for P2(R) and R^3, respectively. Is S an isomorphism between P2(R) and R^3?

Transcript

00:01 Alright, this is a long question.

00:05 So this question is basically about linear transformation and linear function.

00:12 So let's look at question 1.

00:15 So s is a set containing this vectors from v1 to vn.

00:22 It is a set of linear dependent vector in a vector space v.

00:27 So from v1 to v .n, they are linear dependent.

00:31 Which means there must be at least one vector, let's call it vi.

00:37 One vector in this set can be expressed as the linear combination of other vectors.

00:45 That is what we call linear dependent.

00:48 So let's say, well, the vector vi in this set can be expressed by linear combination of all the other vectors, alpha -jvj.

01:05 So j equals 1 to n and j is not i.

01:11 J doesn't equal to i.

01:12 So basically all the other vectors, right, we use this way to express it.

01:17 So we sum over alpha j vj from j equals 1 to n, but we need to jump over the v i.

01:28 Right, because we are using this formula to express v i.

01:32 Alright, now we have this.

01:35 And we also know t is a linear transformation.

01:38 So what is a linear transformation? that's a right definition here.

01:42 So linear transformation need to satisfy two properties.

01:46 So t x plus y should equal to tx plus ty.

01:51 And also t alpha x, alpha is a scalar, should equal to alpha tx.

02:00 Right? this is a definition of linear transformation.

02:03 So let's see what is t v i? t v i equals to we plus this into the expression so t sum over j doesn't equal to i alpha j vj and we use this property because the t is a linear transformation so we could put this sum outside the bracket and use the second property we could move the scalar out of the bracket alpha j t vj now we basically you finish the proof, right? because the vector vti is a vector in the new set, right? can be written as a combination of t vj, right? this element can be written as a linear combination of these elements.

03:03 So this set is also linear dependent.

03:11 Okay, this is the first question.

03:14 And the second question, so here we say, well, define the t, the t, from m m by m matrix to real number r.

03:25 So t a equals to a11 plus a12 plus a plus a n.

03:33 And here it says the trace of a.

03:35 So if the t is a trace of a, here is not a12, but it should be a22, right? because the trace is the sum of the diagonal element.

03:47 In this expression it looks like the sum of all the element in the matrix but it doesn't really matter because they're all linear transformation.

03:55 So we don't, let's do not focus on the detail if it is a trace or not.

04:02 But in any case, we could use here the definition of linear transformation, right? we need to verify both of the property.

04:12 So we want to prove ta plus tb equals to d a plus b.

04:23 Well, it's better to write in this way.

04:26 T -a -plus -b equals to t -a -plus -p -b.

04:31 And this is straightforward because the new matrix a -b, sorry, a -plus -b, should have the element a -1 -1 -1 -2 plus b -1 -a -1 -2, and in the end, a -n -n -n -n -p -n, right? that is a matrix addition, is the addition of the corresponding element...

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