- #1

- 74

- 1

## Homework Statement

Let V = Span{(1,1,0), (1,2,3)}. Define a linear transformation L: V => R^3 by L(1,1,0) = (1,0,0) and L(1,2,3) = (0,1,0). For any (x,y,z) element of V find L(x,y,z)

## Homework Equations

## The Attempt at a Solution

It seems like there should be some straightforward approach to finding the matrix for the transformation. But I keep getting lost. So the standard thm is :

Let L be a linear transformation from V to W and let

S = {v1, ... ,vm} and T = {w1, ... ,wn}

be bases for V and W respectively. Then the matrix A representing L with respect to the bases S and T has ijth component that is the jth coordinate of the vector L(vi).

But then L(1,1,0) = (1,0,0) = 1e

_{1}+ 0e

_{2}+ 0e

_{3}and

L(1,2,3) = (0,1,0) = 0e

_{1}+ 1e

_{2}+ 0e

_{3}(where the e

_{i}'s are the standard bases). But then I just have

1 0

0 1

0 0

which won't even multiply with a 3 term vector.

Another approach I tried and I think worked is just to note any element of V will look like a(1,1,0) + b(1,2,3). Then L( a(1,1,0) + b(1,2,3) ) = aL(1,1,0) + bL(1,2,3) = a(1,0,0) + b(0,1,0). So for any x,y,z I just need to find the corresponding a and b. So any x,y,z element of V must have the relationship that a + b = x, a + 2b = y, and 3b = z for some a,b. This last equation implies b = z/3. Next I just need to find a: a + b = x -> a = x - b and the previous result of b = z/3 -> a = x - z/3. So then I want to say that for any x,y,z element of V. L maps it to (x - z/3, z/3, 0). Does this work? Really, I just want to figure out the direct approach.