Thread: Linear Algebra please help.
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Old June 20th, 2008, 02:13 AM
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Isomorphism Isomorphism is offline
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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.
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