How to show a subspace must be all of a vector space

[GlassBones]

Homework Statement

Show that the only subspaces of $V = R^2$ are the zero subspace, $R^2$ itself,
and the lines through the origin. (Hint: Show that if W is a subspace of
$R^2$ that contains two nonzero vectors lying along different lines through
the origin, then W must be all of $R^2$)

Homework Equations

The Attempt at a Solution

Suppose W is a subspace of $R^2$ that contains two nonzero vectors lying along different lines through the origin ( $x,y \in W,$ $x,y≠0, cx\neq y$ and $cy\neq x$)

Not sure how to proceed to show W must be all of $R^2$. Intuitively this makes sense since I can take a linear combination of vectors in W that should get me to any point in $R^2$. I don't know how to make a formal proof.

Side question: If i have nonzero vectors, is it implied that it goes through the origin?

 

[member 587159]

I suggest another approach. Let $V$ be a subspace of $\mathbb{R}^2$. What can you tell about the possible dimensions? What does each dimension correspond too?

On another note, that the two vectors are not on the same line means that they are linearly independent, so they span a 2 dimensional space.

 

[GlassBones]

The possible dimensions in V are 1,2? 1 corresponds to a point and 2 corresponds to a line.

 

[PeroK]

Homework Statement

Show that the only subspaces of $V = R^2$ are the zero subspace, $R^2$ itself,
and the lines through the origin. (Hint: Show that if W is a subspace of
$R^2$ that contains two nonzero vectors lying along different lines through
the origin, then W must be all of $R^2$)

Homework Equations

The Attempt at a Solution

Suppose W is a subspace of $R^2$ that contains two nonzero vectors lying along different lines through the origin ( $x,y \in W,$ $x,y≠0, cx\neq y$ and $cy\neq x$)

Not sure how to proceed to show W must be all of $R^2$. Intuitively this makes sense since I can take a linear combination of vectors in W that should get me to any point in $R^2$. I don't know how to make a formal proof.

Side question: If i have nonzero vectors, is it implied that it goes through the origin?

All vectors in a vector space start at the origin, if you want to put it that way.

In terms of a proof, you could show that every vector in $\mathbb{R^2}$ is a linear combination of your two vectors.

PS I'm assuming you haven't covered bases and dimensions yet.

 

[GlassBones]

PS I'm assuming you haven't covered bases and dimensions yet.

This is correct.

 

[member 587159]

The possible dimensions in V are 1,2? 1 corresponds to a point and 2 corresponds to a line.

False. 0,1,2

0 corresponds to the null space
1 to a line
2 to the entire plane

 

[PeroK]

This is correct.

Okay, so go with my suggestion above.

 

[GlassBones]

On another note, that the two vectors are not on the same line means that they are linearly independent, so they span a 2 dimensional space.

In terms of a proof, you could show that every vector in R2 is a linear combination of your two vectors.

Okay this seems to be the most straight forward way, I think I was over complicating things.

 

[Mark44]

0 corresponds to the null space

It's probably clearer to say that a dimension of 0 corresponds to the origin-- (0, 0) in $\mathbb R^2$, or the zero vector.