A problem on linear transformation and standard basis

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Main Question or Discussion Point

Problem

Given a transformation T : P(t) -> (2t + 1)P(t) where P(t) ϵ P3
(a) Show that transformation is linear.
(b) Find the image of P(t) = 2 t^2 - 3 t^3
(c) Find the matrix of T relative to the standard basis ε = {1, t, t^2, t^3}
(d) Find the matrix of T relative to the basis β1 = {1, (1-t), (1-t)^2, (1-t)^3}
(e) Suppose [x]β1 = a column vector of 5 rows in which all entries = 2
find [x]ε

Attempt
For part a, should I show that the transformation of P(t) + transformation of Q(t) = transformation of P(t) + Q(t) ?
For part b, do I simply multiply P(t) by (2t+1) ?
I have no idea about parts c, d, e

Note: This is not a homework problem. In fact, I finished schooling years back. I need answers to these, with explanations please, to help somebody else. This thing was never in my syllabus. I had matrix but nothing related to basis.
So, could you explain to me or refer me to some web page from where I can learn this stuff? Thank you.
 

Answers and Replies

  • #2
Deveno
Science Advisor
906
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Problem

Given a transformation T : P(t) -> (2t + 1)P(t) where P(t) ϵ P3
(a) Show that transformation is linear.
(b) Find the image of P(t) = 2 t^2 - 3 t^3
(c) Find the matrix of T relative to the standard basis ε = {1, t, t^2, t^3}
(d) Find the matrix of T relative to the basis β1 = {1, (1-t), (1-t)^2, (1-t)^3}
(e) Suppose [x]β1 = a column vector of 5 rows in which all entries = 2
find [x]ε

Attempt
For part a, should I show that the transformation of P(t) + transformation of Q(t) = transformation of P(t) + Q(t) ?
For part b, do I simply multiply P(t) by (2t+1) ?
I have no idea about parts c, d, e

Note: This is not a homework problem. In fact, I finished schooling years back. I need answers to these, with explanations please, to help somebody else. This thing was never in my syllabus. I had matrix but nothing related to basis.
So, could you explain to me or refer me to some web page from where I can learn this stuff? Thank you.
for part (a) that's PART of what you need to do. you also need to show that if you multiply P(t) by a scalar c, that T(cP(t)) = c(T(P(t)).

for part (b), yes, you apply T to the given polynomial.

(c) actually, a matrix depends on TWO bases, that of the domain, and that of the co-domain (often called "the range" although that is not 100% accurate).

T:P3 → P4, i would assume you will use the basis {1,t,t2,t3,t4} for P4.

the columns of the matrix for T will be the coefficients of T(1), T(t), T(t2), and T(t3) (constant first, in increasing order of powers of t, since that is how the basis is listed).

(d) will take a long time to answer, especially if you are shaky on what a basis is. but the general idea is this:

we define a "change of basis" matrix P that converts β1-coordinates to "standard basis coordinates". then you apply the matrix of part (c), let's call it M, then another matrix that changes ε-coordinates back to β1-coordinates (i use these terms loosely, because your not really making it clear what the 2nd basis is for the image space of T, you're not specifying ANY basis for P4, and you really need to).

QMP, which will be the matrix for T in the new basis (β1-coordinates to β1-coordinates).

for part (e) use P to change from β1-coordinates to ε-coordinates (standard coordinates).
 
  • #3
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@Deveno,
Thank you. So, are the following correct?
a) T(P(t)) = (2t+1)P(t)
T(Q(t)) = (2t+1)Q(t)
T(P(t) + Q(t)) = (2t+1)(P(t) + Q(t))
T(P(t) + Q(t)) = (2t+1)P(t) + (2t+1)Q(t)
T(P(t) + Q(t)) = T(P(t)) + T(Q(t))----------(1)

Also, T(cP(t)) = (2t+1) (c P(t)) = c (2t+1) P(t)
T(cP(t)) = c T(P(t)) ------------(2)

From (1) and (2), the transformation T is linear.

(b) T(P(t)) = (2t+1)P(t)
= (2t+1) (2 t^2 - 3 t^3)
= 4t^3 - 6t^4 + 2t^2 - 3t^3
= -6t^4 + t^3 + 2t^2

(c)
T(1) = 2t+1 = 1 + 2t + 0t^2 + 0t^3 + 0t^4
T(t) = (2t+1)(t) = 2t^2 + t = 0 + t + 2t^2 + 0t^3 + 0t^4
T(t^2) = (2t+1) (t^2) = 2t^3 + t^2 = 0 + 0t + t^2 + 2t^3 + 0t^4
T(t^3) = (2t+1)(t^3) = 2t^4 + t^3 = 0 + 0t + 0t^2 + t^3 + 2t^4

The matrix =
1 ... 0 ... 0 ... 0
2 ... 1 ... 0 ... 0
0 ... 2 ... 1 ... 0
0 ... 0 ... 2 ... 1
0 ... 0 ... 0 ... 2

Could you clarify answers of parts d and e more? I have posted the question in part d exactly as given.

Part (e) says
Suppose [x]β1 =
2
2
2
2
find [x]ε
Earlier I had said 5 rows. I just now realized there are 4 rows.
 
  • #4
Deveno
Science Advisor
906
6
the thing is, it makes no sense to just specify one basis for a matrix.

a matrix is just numbers (field entries), we need the bases to tell us what those numbers stand for.

β1 is a basis for P3, what basis are we supposed to express the column entries of the matrix for T in? is it the standard basis, or {1,1-t,(1-t)2,(1-t)3,(1-t)4} do you see the problem? we'll get "different coordinates" depending on which basis we use (and a different matrix for T).

if your book doesn't say, they've left a very important piece of information out.
 
  • #5
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@Deveno, the book does not say. I typed the complete question above. So, could you answer based on what could we most likely ?
Thank you
 
  • #6
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What if the basis for P4 is given? What if it is {1, t, t^2, t^3, t^4} ?
Then how do we answer?
 
  • #7
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What if the basis for P4 is given? What if it is {1, t, t^2, t^3, t^4} ?
Then how do we answer?
 
  • #8
Deveno
Science Advisor
906
6
if the basis for P4 is just the standard one, then you only need to calculate

the change of basis matrix P, which is:

[tex]\begin{bmatrix}1&1&1&1\\0&-1&-2&-3\\0&0&1&3\\0&0&0&-1\end{bmatrix}[/tex]

and then calculate MP (where M is the matrix for T you already found).

P changes β1-coordinate vectors to their ε-coordinate form:

P(1) = P([1,0,0,0]β1) = [1,0,0,0]ε = 1+0t+0t2+t3

P((1-t)) = P([0,1,0,0]β1) = [1,-1,0,0]ε = 1-t+0t2+0t3

P((1-t)2) = P([0,0,1,0]β1) = [1,-2,-1,0]ε = 1-2t+t2+0t3

P((1-t)3) = P([0,0,0,1]β1) = [1,-3,3,-1]ε = 1-3t+3t2-t3
 

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