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

[JD_PM]

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!

 

[Mark44]

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.

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.

 

[PeroK]

There's a simple way to show that $U \subseteq W$.

Proving that $W$ is not 3D should be simple as well.

 

[FactChecker]

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.

 

[JD_PM]

Thank you all for the helpful replies! :)

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

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

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 $W$.

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

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

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?

 

[PeroK]

There are no proper 3D subspaces. If a subspace is 3D then it's the entire space.

 

[JD_PM]

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.

 

[Mark44]

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.

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.

 

[FactChecker]

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.

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.)