Given subspaces ##U \& W##, show they are equal | Linear Algebra

In summary: I see. You are absolutely right. Thanks for the help! :)In summary, to show that ##U = span \{ (1, 2, 3), (-1, 2, 9)\}## and ##W = \{ (x, y, z) \in \Bbb R^3 | z-3y +3x = 0\}## are equal, we can determine the dimension of each subspace separately and then use the fact that ##dim U = dim W \iff U = W##. To find the dimension of ##U##, we can construct a ##2 \times 3## matrix and use Gaussian elimination, which yields no zero rows and therefore a
  • #1
JD_PM
1,131
158
Homework Statement
Show that ##U = span \{ (1, 2, 3), (-1, 2, 9)\}## and ##W = \{ (x, y, z) \in \Bbb R^3 | z-3y +3x = 0\}## are equal.
Relevant Equations
N/A
Show that ##U = span \{ (1, 2, 3), (-1, 2, 9)\}## and ##W = \{ (x, y, z) \in \Bbb R^3 | z-3y +3x = 0\}## are equal.

I have the following strategy in mind: determine the dimension of subspaces ##U## and ##W## separately and then make use of the fact ##dim U = dim W \iff U=W##. For ##U## I would construct a ##2 \times 3## matrix and use Gaussian elimination on it. My issues are

1) Is the dimension of the row space the one of interest? If that is the case, one gets that ##dim U = 2## because the Gaussian elimination yields no zero rows.

2) For ##W## I do not see how to get the dimension, because I am not given the basis elements of ##W##.Your guidance is appreciated.

Thanks! :biggrin:
 
Physics news on Phys.org
  • #2
JD_PM said:
Homework Statement:: Show that ##U = span \{ (1, 2, 3), (-1, 2, 9)\}## and ##W = \{ (x, y, z) \in \Bbb R^3 | z-3y +3x = 0\}## are equal.
Relevant Equations:: N/A

Show that ##U = span \{ (1, 2, 3), (-1, 2, 9)\}## and ##W = \{ (x, y, z) \in \Bbb R^3 | z-3y +3x = 0\}## are equal.

I have the following strategy in mind: determine the dimension of subspaces ##U## and ##W## separately and then make use of the fact ##dim U = dim W \iff U=W##. For ##U## I would construct a ##2 \times 3## matrix and use Gaussian elimination on it. My issues are

1) Is the dimension of the row space the one of interest? If that is the case, one gets that ##dim U = 2## because the Gaussian elimination yields no zero rows.
It should be obvious that the dimension of U is 2, since the two given vectors span U, and are clearly linearly independent (neither is a multiple of the other). I don't see that you need to be thinking about row space here.
JD_PM said:
2) For ##W## I do not see how to get the dimension, because I am not given the basis elements of ##W##.
For W, you have a single equation in three variables, so there are two degrees of freedom; i.e., the solution space is a plane in ##\mathbb R^3##.

My advice is to find a basis for the W subspace, and then show that it is also a basis for U.
The equation that defines W is z - 3y + 3x = 0, or equivalently,
z = -3x + 3y, where x and y are arbitrary. It doesn't matter which two variables you consider as arbitrary -- once you set anyone variable, the other two variables can be chosen arbitrarily.

From the equation above, you can write this system of equations.
x = x
y = . . . . . y
z = -3x + 3y

A basis fairly leaps out at you from the system above.
 
  • Like
Likes JD_PM
  • #3
There's a simple way to show that ##U \subseteq W##.

Proving that ##W## is not 3D should be simple as well.
 
  • Like
Likes FactChecker
  • #4
Can you show that the two vectors in U are also in W? That would probe that dim(W) ##\ge## dim(U) = 2. Then can you find a vector that is not in W? That would prove that dim(W) ##\lt## dim(##\mathbb{R}^3##) = 3.
 
  • Like
Likes JD_PM and PeroK
  • #5
Thank you all for the helpful replies! :)

Mark44 said:
A basis fairly leaps out at you from the system above.

Indeed! I now see that a possible basis for ##W## is ##\beta = \{ (1, 1, 0), (0, 1, 3)\}##.

FactChecker said:
Can you show that the two vectors in U are also in W? That would probe that dim(W) ##\ge## dim(U) = 2.

That can shown by writing these as a linear combination of the basis vectors of ##W##.

$$(1,2,3) = \underbrace{k_1}_{=1}(1,1,0) + \underbrace{k_2}_{=1}(0,1,3) $$

$$(-1,2,9) = \underbrace{k_1}_{=-1}(1,1,0) + \underbrace{k_2}_{=3}(0,1,3) $$

FactChecker said:
Then can you find a vector that is not in W? That would prove that dim(W) ##\lt## dim(##\mathbb{R}^3##) = 3.

Yes, say ##(1,1,1)## because we cannot write it as a linear combination of the basis vectors of ##W##.

However, why does that prove that dim(W) ##\lt## dim(##\mathbb{R}^3##) = 3?
 
  • #6
There are no proper 3D subspaces. If a subspace is 3D then it's the entire space.
 
  • Like
Likes JD_PM
  • #7
PeroK said:
There are no proper 3D subspaces. If a subspace is 3D then it's the entire space.

Ahhh so the fact that we can find a vector ## v \in (x, y, z)## that is not in ##W## means that ##dim(W) < dim( \Bbb R^3) = 3## We would find any ## v \in (x, y, z)## in ##W## otherwise.
 
  • #8
Mark44 said:
From the equation above, you can write this system of equations.
x = x
y = . . . . . y
z = -3x + 3y

A basis fairly leaps out at you from the system above.

JD_PM said:
Indeed! I now see that a possible basis for ##W## is ##\beta = \{ (1, 1, 0), (0, 1, 3)\}##.
Maybe the first vector works, but I didn't check. The basis that "leaps out" from my work is {<1, 0, -3>, <0, 1, 3>}.

More explicitly, the system of equations I wrote is
x = 1x + 0y
y = 0x + 1y
z = -3x + 3y
The coefficients of x are the first basis vector, and the coefficients of y are the second one.
 
  • Like
Likes JD_PM
  • #9
JD_PM said:
That can shown by writing these as a linear combination of the basis vectors of ##W##.

$$(1,2,3) = \underbrace{k_1}_{=1}(1,1,0) + \underbrace{k_2}_{=1}(0,1,3) $$

$$(-1,2,9) = \underbrace{k_1}_{=-1}(1,1,0) + \underbrace{k_2}_{=3}(0,1,3) $$
I would prefer that you just plug the numbers into the original equation defining W and show that the equation is satisfied.
JD_PM said:
Yes, say ##(1,1,1)## because we cannot write it as a linear combination of the basis vectors of ##W##.

However, why does that prove that dim(W) ##\lt## dim(##\mathbb{R}^3##) = 3?
Again, I would prefer that you plug the numbers into the original equation defining W and show that it is not satisfied. Since the subspace generated by (1,1,1) is not contained in W, the dimension of W must be less than 3. (I am not sure what theorems are available for you to use to formally prove that.)
 
  • Like
Likes JD_PM and PeroK

What does it mean for two subspaces to be equal?

Two subspaces are considered equal if they contain the same set of vectors. This means that every vector in one subspace is also in the other subspace, and vice versa.

How can I show that two subspaces are equal?

To show that two subspaces are equal, you can prove that one is a subset of the other and vice versa. This can be done by showing that every vector in one subspace can be written as a linear combination of vectors in the other subspace.

What is the importance of proving that two subspaces are equal?

Proving that two subspaces are equal is important in understanding the relationship between different subspaces. It also helps in solving problems involving linear transformations and finding bases for subspaces.

What are some common methods for proving that two subspaces are equal?

Some common methods for proving that two subspaces are equal include showing that they have the same dimension, showing that they have the same basis, and showing that they have the same span.

Are there any cases where two subspaces can be equal without having the same basis or dimension?

Yes, it is possible for two subspaces to be equal without having the same basis or dimension. This can happen if one subspace is a subset of the other and they share some common vectors, or if one subspace is a linear combination of vectors in the other subspace.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
589
  • Calculus and Beyond Homework Help
Replies
14
Views
570
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
439
  • Calculus and Beyond Homework Help
Replies
15
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
597
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
5
Views
5K
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
Back
Top