Polynomial Basis and Linear Transformation

[zairizain]

Homework Statement

Let X be the vector space of polynomial of order less than or equal to M

a) Show that the set B={1,x,...,x^M} is a basis vector

b) Consider the mapping T from X to X defined as:

f(x)= Tg(x) = d/dx g(x)

i) Show T is linear

ii) derive a matrix representation for T in terms of the basis B

iii) what are the eigenvalues of T

iv) compute one eigenvector associated with one of the eigenvalues

Homework Equations

The Attempt at a Solution

a) i)Linear independence;

a1(1) + a2(x)+...+an(x^M) = 0

a1=a2=an=0

ii)Span

a+bx+...+cx^M=0

Such that; a1(1) +a2(x)+...+an(x^M) = a+bx+...+cx^M

a1=a, a2=b, an=c


b)

i) f(x) = a0 + a1X+...+amX^M
g(x) = b0 + b1X+...bmX^M
g(t) = b0+b1t+...+bmt^M

Tg(t) = b0t + b1t^2+...+bmt^(M+1)

For any scalar, k is element K

T(k g(t)) = t (k g(t))
= k (t g(t))
= KT (g (t))

Thus T is linear


ii) B= {1, x ,x^2,...,x^M}

matrix T,=

0 0 0 ...0
0 1 0 ...0
0 0 2 ...0
0 0 0 ...0
. . . .. .
0 0 0 .. M


iii) eigenvalues of T, lambda = [0, 1, 2...M]

iv)for lambda = 1;

(A-lambda I)=0
(A- I ) = 0


[matrix T] [ a1;a2;...am] = [ a1;a2;...am]

a1=a2=...=am

eigenvector for lambda =1 is;

[1, 1, ...1]


Is this correct?Please help me.

 

[Mark44]

Homework Statement

Let X be the vector space of polynomial of order less than or equal to M

a) Show that the set B={1,x,...,x^M} is a basis vector

This should be "Show that the set B={1,x,...,x^M} is a basis for X."

b) Consider the mapping T from X to X defined as:

f(x)= Tg(x) = d/dx g(x)

i) Show T is linear

ii) derive a matrix representation for T in terms of the basis B

iii) what are the eigenvalues of T

iv) compute one eigenvector associated with one of the eigenvalues

Homework Equations

The Attempt at a Solution

a) i)Linear independence;

a1(1) + a2(x)+...+an(x^M) = 0

It's probably more convenient to label the constants as a0, a1, a2, ..., aM. That way they match the exponent on x.

a1=a2=an=0

This is true whether the functions are linearly independent or linearly dependent. The critical difference is whether this is the unique solution (independent set of functions) or one of many (dependent set).

How can you establish that there are no other solutions for a0, a1, a2, ..., aM?

ii)Span

a+bx+...+cx^M=0

Such that; a1(1) +a2(x)+...+an(x^M) = a+bx+...+cx^M

a1=a, a2=b, an=c

Why do you think that a+bx+...+cx^M=0? Are there only three terms on the left side? That's what you're implying with constants a, b, and c.

When you say "Such that; a1(1) +a2(x)+...+an(x^M) = a+bx+...+cx^M" what is a3? a4?

b)

i) f(x) = a0 + a1X+...+amX^M
g(x) = b0 + b1X+...bmX^M
g(t) = b0+b1t+...+bmt^M

No, this is wrong. f(x) = d/dx(g(x)). There is no t, but there is T and these are different letters.

For example, if g(x) = b0 + b1x + b2x + ... + bMx^M, what is f(x)?

Tg(t) = b0t + b1t^2+...+bmt^(M+1)

For any scalar, k is element K

T(k g(t)) = t (k g(t))
= k (t g(t))
= KT (g (t))

Thus T is linear


ii) B= {1, x ,x^2,...,x^M}

matrix T,=

0 0 0 ...0
0 1 0 ...0
0 0 2 ...0
0 0 0 ...0
. . . .. .
0 0 0 .. M


iii) eigenvalues of T, lambda = [0, 1, 2...M]

iv)for lambda = 1;

(A-lambda I)=0
(A- I ) = 0


[matrix T] [ a1;a2;...am] = [ a1;a2;...am]

a1=a2=...=am

eigenvector for lambda =1 is;

[1, 1, ...1]


Is this correct?Please help me.