• Jun 19th 2008, 09:51 PM
JCIR
T:R2-->R3 defined by T(a1,a2)=(a1+a2,0,2a1-a2)

I need to prove that is a linear transformation, and find the bases for both
N(T) and R(T). then compute the nullity and rank of T. finally say where it is one -2- one or onto.

I already got the anwsers but I need to know in detail especially for computing the nullity and rank of T. and onto or one 2 one.
Thanks
• Jun 20th 2008, 12:13 AM
Isomorphism
Quote:

Originally Posted by JCIR
T:R2-->R3 defined by T(a1,a2)=(a1+a2,0,2a1-a2)

I need to prove that is a linear transformation

Its just checking rules right? Do it.

Or get a matrix that represents the transformation.

$\displaystyle T \equiv \begin{pmatrix}1 & 1 \\ 0 & 0 \\ 2 & -1 \end{pmatrix}$

Quote:

and find the bases for both
Consider the standard base for $\displaystyle \mathbb{R}^2 = \{(0,1),(1,0)\}$.

Now $\displaystyle T(0,1) = (1,0,-1),T(1,0) = (1,0,2)$

So the basis for $\displaystyle T(\mathbb{R}^2)$.

Quote:

N(T) and R(T). then compute the nullity and rank of T. finally say where it is one -2- one or onto.
N(T) is $\displaystyle (x,y):T(x,y) = (x+y,0,2x - y) = 0 \Rightarrow -x = y = 2x \Rightarrow (x,y) = 0.$.This means Nullity = 0

R(T) is $\displaystyle (a,b,c):T(x,y) = (x+y,0,2x - y) = (a,b,c) \Rightarrow y = a-x, 2x-y = 3x - a = c$$\displaystyle \Rightarrow (x,y) = (\frac{a+c}3,\frac{2a- c}3 ).$.

This means R(T) will be the subspace of R3, in which the second co-ordinate is 0.This means Rank = 2.

The map is one-one because nullity is 0.