1. ## Linear Transformations

Define T: P3-->P3 by T(f)(t)=2f(t)+(1-t)f'(t)
Show that T is a linear transformation.

I've included the problem that I'm working on. I'm having a hard time proving this transformation is linear because of the (1-t)f'(t) term at the end of it. I'm not entirely sure what to do with it when I evaluate T(s+t)=T(s)+T(t). Do I make it (1-t+s)f'(t)?

Also looking ahead, I'm not sure how to approach part b) as he skipped that in lecture.

2. ## Re: Linear Transformations

Originally Posted by renolovexoxo

Define T: P3-->P3 by T(f)(t)=2f(t)+(1-t)f'(t)
Show that T is a linear transformation.
Use better notation.
$\mathcal{T}(f(t))=2f(t)+(1-t)f'(t)$, $\mathcal{T}(g(t))=2g(t)+(1-t)g'(t)$

Does that sum equal $\mathcal{T}[f+g](t))=2[f+g](t)+(1-t)[f+g]'(t)~?$

3. ## Re: Linear Transformations

Thanks so much! Any help on part b)?

4. ## Re: Linear Transformations

for part (b) recall that [T]B where B is a basis B = {v1,...,vn}}, has columns [T(vj)]B.

therefore, find T(1), T(t), T(t2) and T(t3), and express these as coordinates in the basis given.

5. ## Re: Linear Transformations

For part b) that was how i started the first few times i attempted this problem. I'm just not sure what to do with the T(t) term that equals 2t+(1-t) and the T(t^3)=2t^3+2t^2-3t^5. I put in in like this, but I think its not correct, as I'm not sure what happens with the (1-t) term and the t^5 term in this situation:

[2 (1-t) 0 0
0 2 -2 0
0 0 2 3
0 0 -2 2]

6. ## Re: Linear Transformations

Originally Posted by renolovexoxo
For part b) that was how i started the first few times i attempted this problem. I'm just not sure what to do with the T(t) term that equals 2t+(1-t) and the T(t^3)=2t^3+2t^2-3t^5. I put in in like this, but I think its not correct, as I'm not sure what happens with the (1-t) term and the t^5 term in this situation:

[2 (1-t) 0 0
0 2 -2 0
0 0 2 3
0 0 -2 2]
no, that is nowhere near correct. you are not evaluating T correctly. we can write an element of P3 as: f(t) = a + bt + ct2 + dt3

(P3 is a 4-dimensional space, and in the basis B = {1,t,t2,t3}, we have [f]B = [a,b,c,d]B).

using the definition of T, we have: T(f(t)) = 2f(t) + (1-t)f'(t). thus:

T(f(t)) = T(a + bt + ct2+dt3) = 2(a + bt + ct2 + dt3) + (1 - t)(b + 2ct + 3dt2)

= 2a + 2bt + 2ct2 + 2dt3 + b + 2ct + 3dt2 - bt - 2ct2 - 3dt3

= (2a + b) + (2b + 2c - b)t + (2c + 3d - 2c)t2 + (2d - 3d)t3

= (2a + b) + (b + 2c)t + 3dt2 - 3dt3 <---no powers of t higher than 3.

for the function f(t) = 1 (a constant function), a = 1, b = c = d = 0, so

T(1) = 2.

for the function f(t) = t, we have a = 0, b = 1, c = d = 0, so

T(t) = 1 + t

for the function f(t) = t2, we have a = b = 0, c = 1, d = 0, so:

T(t2) = 2t

for the function f(t) = t3, we have a = b = c = 0, d = 1, so

T(t3) = 3t2 - 3t3

you're thinking of "t" as something that T acts on. it's not. it's what you call a "dummy variable", so that we don't have to write f as something like:

f = (constant1)*(constant function 1) + (constant2)*(identity function) + (constant3)*(squaring function) + (constant4)*(cubing function).

we could easily think of T as the function that goes from R4 to R4:

T(a,b,c,d) = (2a+b,b+2c,3d,-3d), or more properly:

[T]B([a,b,c,d]B) = [2a+b,b+2c,3d,-3d]B

since P3 is isomorphic to R4 using the isomorphism:

(1,0,0,0) → 1
(0,1,0,0) → t
(0,0,1,0) → t2
(0,0,0,1) → t3

in other words, you need to start thinking of a polynomial in t as being "a vector consisting of its coefficients".

7. ## Re: Linear Transformations

What is the solution of C,D and e? Also the answer should be -1t^3 3t^2, not -3t^3.