Prove that T is a linear transformation

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March 12th 2011, 08:40 PM
zodiacbrave

Prove that T is a linear transformation

Hello,

T: P2(R) -> P3(R) defined by T(f(x)) = xf(x) + f`(x).

Let A,B,C,a,b,c be elements of R and x,y elements of P2, where f(x) = (a + bx + cx^2) and f(y) = (A + By + Cy^2)

Can't I write this as a matrix vector product and prove it to be linear that way? I can also calculate the rank and null this way, right?

T(f(x)) = x(a + bx + cx^2) + b + 2x
= b + 2ax + bx^2 + cx^3

is this right so far? I really want to use the matrix vector product method though, but I'm having a hard time making it into a matrix, any advice?
March 12th 2011, 11:02 PM
FernandoRevilla

Quote:

Originally Posted by zodiacbrave

Can't I write this as a matrix vector product and prove it to be linear that way?

Better:

$T[\lambda f(x)+\mu g(x)]=x (\lambda f(x)+\mu g(x))+(\lambda f(x)+\mu g(x))'=$

$\ldots =\lambda T[f(x)]+\mu T[g(x)]$
March 13th 2011, 04:08 AM
HallsofIvy

Fernando Revilla is correct that the best way to prove something is a linear transformation is to show that it satisfies the definition of a linear transformation.

Since you specifically ask about writing it as a matrix, remember that a linear transformation from vector space U to vector space V depends on specific choices of (ordered) bases for U and V. Now, apply the linear transformation to each of the basis vector of U in turn, writing the result in terms of the basis for V. The coefficients will be the columns of the matrix.

Here, U is the function space P2 of quadratic polynomials and V is the function space P3 of cubic polynomials. The standard basis for the first is 1, x, and $x^2$ and for the second, 1, x, $x^2$ , and $x^3$ .

Applying the linear transformation to "1" we get $x(1)+ 0= x= 0(1)+ 1(x)+ 0(x^2)+ 0(x^3)$ . The first column of the matrix representation is $\begin{bmatrix}0 \\ 1\\ 0 \\ 0\end{bmatrix}$ .

Applying the linear transformation to "x" we get $x(x)+ 1= x^2+ 1= 1(1)+ 0(x)+ 1(x^2)+ 0(x^3)$ . The second column of the matrix representation is $\begin{bmatrix}1 \\ 0 \\ 1 \\ 0\end{bmatrix}$ .

Applying the linear transformation to " $x^2$ " we get $x(x^2)+ 2x= x^3+ 2x= 0(1)+ 2(x)+ 0(x^2)+ 1(x^3)$ . The third column of the matrix representation is $\begin{bmatrix}0 \\ 2 \\ 0 \\ 1\end{bmatrix}$ .

The matrix representing the linear transformation, using these bases in these orders, is
$\begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 2\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$ .