• Jun 19th 2008, 10:01 PM
JCIR
Let T: P2(R)--->P3(R) be a linear transformation defined by

T(f(x))= 2f'(x) + Integral from 0 to x of 3f(t)dt.

i know R(T)= span( T(1), T(x), T(x^2) ) = Span( (3x, 2 + 3/2x^2,4x+x^3)

But what i need to know is now to they go from T(1) to 3x

from T(x) to ....

Also if possible if you can tell me how to find the rank, nullity, and if is one-to one or onto will be a great help.

thanks alot.
• Jun 19th 2008, 11:55 PM
Isomorphism
Quote:

Originally Posted by JCIR
Let T: P2(R)--->P3(R) be a linear transformation defined by

T(f(x))= 2f'(x) + Integral from 0 to x of 3f(t)dt.

i know R(T)= span( T(1), T(x), T(x^2) ) = Span( (3x, 2 + 3/2x^2,4x+x^3)

But what i need to know is now to they go from T(1) to 3x

from T(x) to ....

Also if possible if you can tell me how to find the rank, nullity, and if is one-to one or onto will be a great help.

thanks alot.

First prove that dim(Span(3x, 2 + 3/2x^2,4x+x^3)) = 3

Its actually simple. Observe that Span(3x, 2 + 3/2x^2,4x+x^3) has a dimension of 3. Observe that the rank is 3.
$\displaystyle a+bx + cx^2$ maps to $\displaystyle 2b + (3a+4b)x + 3b/2 x^2 + c x^3$. Thus matrix of Transformation is $\displaystyle \begin{pmatrix} 0 & 2 & 0 \\3 & 4 & 0 \\0 & 3/2 & 0 \\0 & 0 & 1 \end{pmatrix}$

Its easy to check that the rank of the above matrix is 3.

Hence dimension(image) = rank = 3.
Thus by rank-nullity theorem nullity = 1 and thus the kernel has a dimension 1. This also implies that the Transform is not one-one.