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)

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.

 

[HallsofIvy]

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.

 

[ayh123]

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?

 

[HallsofIvy]

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.

 

[ayh123]

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!