# Which of the following are in R(T)?

• ayh123

## Homework Statement

Let T: R^4 ---> R^3 be the linear transformation given by the formula:
T(x1, x2, x3, x4) = (4x1+x2-2x3-3x4, 2x1+x2+x3-4x4, 6x1-9x3+9x4)

Which of the following are in R(T)?

(a) (0,0,6)
(b) (1,3,0)
(c) (2,4,1)

## 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.

Essentially you are asked to determine if a set of equations has a solution.

For example, (0, 0, 6) is in R(T) if and only if there exist $(x_1, x_2, x_3, x_4)$ such that $T(x_1, x_2, x_3, x_4)= (4x_1+x_2-2x_3-3x_4, 2x_1+x_2+x_3-4x_4, 6x_1-9x_3+9x_4)= (0, 0, 6)$. In other words, if and only if the equations
$4x_1+ x_2- 2x_3- 3x_4= 0$
$2x_1+ x_2+ x_3- 4x_4= 0$
$6x_1- 9x_3+ 9x_4= 6$
have at least one solution

Your method, however, of finding a basis for the range space is perfectly good- and faster. I see that you got three vectors for the basis. That means they span all of R3 and so, not just these but any set of three numbers, (a, b, c), is in that range.

Of course, a single vector space has many different bases. The fact that you happen to get a different one from what the book gives is not a problem.

Thanks for the clarification. Is this a common range for T - the set of all real numbers? Is there any way to gauge if that is the case without row reducing a matrix?

The range of T, here, is NOT "the set of all real numbers". It is all of R3. T here is from R4 to R3.

In general, R(T), where T is from an n dimensional space to an m dimensional space is a subspace of dimension at most n. It is also true that dimension of R(T) plus the dimension of the kernel of T (the "rank" and "nullity" of T, respectively) must be equal to n. Here, T is from a 4 dimensional space to a 3 dimensional space so the rank is at most 3 and nullity at least 1. But it is possible for the rank to be smaller than that.

Thanks. I knew real numbers wasn't right.. but couldn't think of a way to describe R^3.. set of real vectors in three dimensions? But I think the correct way to describe it is the set of all ordered triples of real numbers.

Wanted to clarify one more thing. You said, "In general, R(T), where T is from an n dimensional space to an m dimensional space is a subspace of dimension at most n." Do you mean "at most m"?

Thank you very much for your help!