Finding the Matrix Representation of a Linear Transformation with Given Basis

[mlarson9000]

Homework Statement

Let T: P3-P3 be the linear transformation defined by T(p(x))= D^2(p(x))-4D(p(x)) + p(x). Find the Matrix representation of A of T, where B = (x, 1+x, x+x^2, x^3).

Homework Equations

The Attempt at a Solution

I don't know where to start here. What is D? Is itjust a variable? What is p(x)? is it just a generic funtion? Will the Matrix be filled with D's and p(x)'s? Someone please let me know where to start.

 

[Mark44]

D is the (first) derivative operator. D² is the second derivative operator. p(x) is an arbitrary polynomial in P₃, meaning that it is a polynomial of degree less than or equal to 3. One example is p(x) = 1 + 2x + 3x² + x³.

A polynomial in P₃ can be represented as an ordered 4-tuple, by listing the coefficients of each power of x. The function p in my example can be written in this way as [1 2 3 1]. Your matrix will have numbers in it, not Ds or p(x)'s.

 

[mlarson9000]

So is that the same thing as f"(x)-4f'(x)+f(x)?

 

[Dick]

If you mean T(f(x))=f"(x)-4f'(x)+f(x). Yes, it is. If you want to simply the problem a little, you could find the matrix M of S(f(x))=f'(x). The matrix you are looking for is then is M^2-4M+I.

 

[steve89]

Kind of like Laplace transformation ?
T (f) = S²L(f)-Sf(0)-f^'(0)-4SL^'(f)-4f(0)+f

 

[Mark44]

Kind of like Laplace transformation ?
T (f) = S²L(f)-Sf(0)-f^'(0)-4SL^'(f)-4f(0)+f

I don't see that there's any connection.

 

[HallsofIvy]

Now that you know what the linear transformation is, you can find its matrix representation in a given basis by applying it to each basis "vector" in turn, writing the result as a linear combination of the basis vectors. The coefficients form the columns of the matrix.

For example, your given basis is (x, 1+x, x+x^2, x^3). T(x)= D^2(x)-4D(x) + x= 0- 4(1)+ 1= x- 4= 5(x)- 4(x+1)+ 0(x+x^2)+ 0(x^3). The first column of the matrix is (5, -4, 0, 0).

 

[mlarson9000]

For example, your given basis is (x, 1+x, x+x^2, x^3). T(x)= D^2(x)-4D(x) + x= 0- 4(1)+ 1= x- 4= 5(x)- 4(x+1)+ 0(x+x^2)+ 0(x^3). The first column of the matrix is (5, -4, 0, 0).

So, for the second column: T(1+x)=D^2(1+x)-4D(1+x)+(1+x)= 0-4(1)+(1+x)=x-3=4(x)-3(x+1)+0(x+x^2)+0(x^3) gives <4,-3,0,0>

Right?

 

[HallsofIvy]

Yes! Very good.