Linear Algebra please help. Exam in the morning.

JCIR · #1 $Old$ June 20th, 2008, 12:01 AM

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.

Isomorphism · #2 $Old$ June 20th, 2008, 01:55 AM

Quote:

Originally Posted by JCIR $View Post$

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.
$a+bx + cx^2$ maps to $2b + (3a+4b)x + 3b/2 x^2 + c x^3$ . Thus matrix of Transformation is $\begin{pmatrix} 0 & 2 & 0 \\3 & 4 & 0 \\0 & 3/2 & 0 \\0 & 0 & 1 \end{pmatrix}$

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.