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Linear transformation

  • Thread starter Benny
  • Start date
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Hi, can someone please check my working for the first part and help me out with the second bit?

Q. Let P_2 be the vector space of all polynomials of degree at most 2. A function [tex]T:P_2 \to R^3 [/tex] is defined by [tex]T\left( {p\left( x \right)} \right) = \left( {p(1),p'(1),p''(1)} \right)[/tex]. That is, T evaluates the polynomial p(x) and its first two derivatives at the point x = 1.

a) Show that T is a linear transformation.
b) Find the matrix of T using the basis B = {1,x,x^2} for P_2 and the standard basis S = {(1,0,0),(0,1,0),(0,0,1)} for R^3.

a) Let p(x) and q(x) be elements of P_2 and let c be a scalar.

(i) T(c(p(x))) = T((cp(x)))
= (cp(1),cp'(1),cp''(1))
=c(p(1),p'(1),p''(1))
=cT(p(x))

(ii) T((p(x)) + (q(x)))
= T((p(x) + q(x)))
=((p+q)(1),(p+q)'(1),(p+q)''(1))
=(p(1) + q(1), p'(1) + q'(1), p''(1) + q''(1))
=(p(1),p'(1),p''(1)) + (q(1), q'(1), q''(1))
=T(p(x)) + T(q(x))

b) I don't know how to do this question. I've read through the relevant section of my textbook and there is a part which mentions a linear transformation T: R^n -> R^m and a "unique matrix" A such that Ax = T(x) for all x in R^n. I'm not sure if this is related to the question though since I am not familiar with the terminology of this question so I'm making assumptions as to what they mean - specifically "matrix of T".

Well I guess I'll just list some thoughts. If I use the Ax = T(x) method I would deduce that T is a 3 x 3 matrix. I do this by using coordinates vectors with the basis {1,x,x^2} although it appears to be a rather weak 'guess method.'
I can't think of any other methods that I can use to do this question. I'm going to think about it a bit more but could someone please help me out?
 

HallsofIvy

Science Advisor
Homework Helper
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876
There's a standard way of finding the matrix corresponding to given bases for the domain and range.

Apply the linear transformation to each basis vector for the domain, then write the result in terms of the basis for the range. The coefficients form the columns for the matrix. (The matrix is not unique. It depends upon the ordering of the bases.) In general, a linear transformation from an n dimensional vector space to an m dimensional vector space has n columns and m rows. In this case, yes, it is a 3 by 3 matrix.

For example, applying T to p(x)= x2 gives (1, 2, 4)= 1(1, 0, 0)+ 2(0, 1, 0)+ 4(0, 0, 1). One column is [1 2 4].
 

lurflurf

Homework Helper
2,417
122
Benny said:
b) I don't know how to do this question. I've read through the relevant section of my textbook and there is a part which mentions a linear transformation T: R^n -> R^m and a "unique matrix" A such that Ax = T(x) for all x in R^n. I'm not sure if this is related to the question though since I am not familiar with the terminology of this question so I'm making assumptions as to what they mean - specifically "matrix of T".

Well I guess I'll just list some thoughts. If I use the Ax = T(x) method I would deduce that T is a 3 x 3 matrix. I do this by using coordinates vectors with the basis {1,x,x^2} although it appears to be a rather weak 'guess method.'
I can't think of any other methods that I can use to do this question. I'm going to think about it a bit more but could someone please help me out?
Don't get a linear transorm and a matrix confused. A matix is a box of numbers with an algebra that can be used to repressent a linear transform. What you need to remember is a linear tranform can be defined by its action on a basis. So figure what T does to the basis, then make a matrix that does the same thing.
T(1)=a11(1,0,0)+a12(0,1,)+a13(0,0,1)
T(x)=a21(1,0,0)+a22(0,1,0)+a32(0,0,1)
T(x^2)=a31(1,00)+a32(0,1,0)+a33(0,0,1)
figure the 9 numbers aij they are the same numbers in the same arrangement as the matrix you seek.

Also you part 1 looks good to show a map is linear show
T(cv)=cT(v)
and
T(u+v)=T(u)+T(v)
as you did.
 
584
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Thanks for the help HallsofIvy and lurflurf, much appreciated.
 
584
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I just tried doing this question and I keep on getting matrix A whereas the answer is matrix B.

[tex]
A = \left[ {\begin{array}{*{20}c}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & 2 \\
\end{array}} \right]
[/tex]

[tex]
B = \left[ {\begin{array}{*{20}c}
1 & 1 & 1 \\
0 & 1 & 2 \\
1 & 2 & 2 \\
\end{array}} \right]
[/tex]

Here is what I did(I have skipped the step on writing linear combinations keeping in mind the fact that I am working with R^3).

Basis before transformation: {1,x,x^2}...basis after transformation: R^3.

T(1) = (1,0,0)...P(1) = 1 since p(x) = 1 etc
T(x) = (1,1,0)...P(1) = 1 since p(x) = 1 etc
T(x^2) = (1,2,2)...P(1) = 1 since p(x) = x^2 and then differentiate, substitute..

I formed the matrix A by taking each coordinate vector as the column of A. So (1,0,0) is the column of A. I'm not sure what I'm doing wrong. Can someone please help?
 

lurflurf

Homework Helper
2,417
122
Benny said:
I formed the matrix A by taking each coordinate vector as the column of A. So (1,0,0) is the column of A. I'm not sure what I'm doing wrong. Can someone please help?
Your matrix A is good. The matrix B has different eigen values. I do not know where B came from. B is very odd it may have a typo in the bottome row. Notice that A and B are the same except in the first two columbs of the 3rd row. In particular B implies T(1)=(1,0,1) and T(x)=(1,1,2) which in turn implies
1=(1)''|x=1
2=(x)''|x=1
Which is clearly false.
 
584
0
Oh ok, thanks for your response lurflurf.
 

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