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## Homework Statement

Let T: R^4 ---> R^3 be the linear transformation given by the formula:

T(x1, x2, x3, x4) = (4x

_{1}+x

_{2}-2x

_{3}-3x

_{4}, 2x

_{1}+x

_{2}+x

_{3}-4x

_{4}, 6x

_{1}-9x

_{3}+9x

_{4})

Which of the following are in R(T)?

(a) (0,0,6)

(b) (1,3,0)

(c) (2,4,1)

## Homework Equations

Not sure what this means.. please skip to attempt at a solution

## The Attempt at a Solution

I have tried so many different ways, and it seems to me that all of these vectors are in R(T).

I tried finding the basis of R(T) by putting the coefficients into matrix (A), and row-reducing A to row echelon form, then noting the columns with the leading 1's in them, and matching them with the original A. So then I thought, all vectors in R(T) must be a linear combination of these 3 basis vectors and will satisfy r (4, 2, 6) + s (1, 1, 0) + t (-3, -4, 9) = (b1, b2, b3).

But all of them satisfy these conditions!

I also tried using the elementary vectors, or basis vectors for R^4.

I found the transformation images for:

T(1,0,0,0) = (4, 2, 6)

T(0,1,0,0) = (1, 1, 0)

T(0,0,1,0) = (-2, 1, -9)

T(0,0,0,1) = (-3, -4, 9)

Then I used the proposition that if the vectors span R^4, and it is a linear transformation, then T(vectors) span R(T).

I put the T(vectors) into a matrix and row-reduced. This gave me the 3 linearly independent vectors that are the basis of R(T) (or so I thought).

So then, all vectors in R(T) must be a linear combination of these vectors:

r (1, 1/2, 3/2) + s (0, 1, -3) + t (0, 0, 1) = (b1, b2, b3).

But all of those fit this formula!

A related question in the book actually does ask for the basis of the range, and since it is an odd-numbered problem, in the back, it gives this answer for a basis:

{(1, 0, 0), (0, 1, 0), (3/2, -4, 1)}

which is not what I got in either tries.

I bet you're shaking your head at how wrongly I did this problem. Please help clarify... thanks in advance.