Linear Algebra please help.

JCIR · #1 $Old$ June 19th, 2008, 11:51 PM

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

Isomorphism · #2 $Old$ June 20th, 2008, 02:13 AM

Quote:

Originally Posted by JCIR $View Post$

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.

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

Quote:

and find the bases for both

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

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

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

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 $(x,y):T(x,y) = (x+y,0,2x - y) = 0 \Rightarrow -x = y = 2x \Rightarrow (x,y) = 0.$ .This means Nullity = 0

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

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.