1. ## Subspace problem

The problem states:

Show that the only subspaces of k^2 are the vector space {0} (the zero-vector), a line, and a plane.

Now I am thinking of it as k^2=[x,y]. obviously when [x,y]=[0,0], it is the zero vector, [a,0] would be a line, and [a,b] would be a plane.

However, we are supposed to think of there being a subspace, call it S, where s is not {0} or a line. We should show that S contains two non-collinear vectors and the span of these two vectors should be k^2.

My mind is pretty much blown by this problem and any in all help would be greatly appreciated. Thanks!

2. Originally Posted by sabrepride
The problem states:

Show that the only subspaces of k^2 are the vector space {0} (the zero-vector), a line, and a plane.

Now I am thinking of it as k^2=[x,y]. obviously when [x,y]=[0,0], it is the zero vector, [a,0] would be a line, and [a,b] would be a plane.

However, we are supposed to think of there being a subspace, call it S, where s is not {0} or a line. We should show that S contains two non-collinear vectors and the span of these two vectors should be k^2.

My mind is pretty much blown by this problem and any in all help would be greatly appreciated. Thanks!

Attack the problem directly: suppose $W\subset k^2$ is a subspace, then:

1) If $W$ has one unique element it must be the zero vector and thus $W=\{0\}$

2) If $W$ has at least one non-zero element $w$ look at $:=Span\{w\}$ ; if $W=$ then $W$ is a line (through the origin, of course), otherwise

there exists $v\in W\,,\,\,v\notin =Span\{w\}\Longleftrightarrow \{w,v\}$ is linearly independent and thus $\dim W\geq 2=|\{w,v\}|$ , but of course $\dim W\leq 2=\dim k^2\Longrightarrow \dim W =2$

and $W$ is a plane (again, through the origin)...in fact, then $W=k^2$ which in fact is a plane itself.

Tonio

3. Originally Posted by tonio
Attack the problem directly: suppose $W\subset k^2$ is a subspace, then:

1) If $W$ has one unique element it must be the zero vector and thus $W=\{0\}$

2) If $W$ has at least one non-zero element $w$ look at $:=Span\{w\}$ ; if $W=$ then $W$ is a line (through the origin, of course), otherwise

there exists $v\in W\,,\,\,v\notin =Span\{w\}\Longleftrightarrow \{w,v\}$ is linearly independent and thus $\dim W\geq 2=|\{w,v\}|$ , but of course $\dim W\leq 2=\dim k^2\Longrightarrow \dim W =2$

and $W$ is a plane (again, through the origin)...in fact, then $W=k^2$ which in fact is a plane itself.

Tonio
Thanks for your quick reply. Is there anyway you could speak of this without reference to dimension. We have no covered this is my course yet and thus I would like to solve it the way the professor intends.
Also, what do you mean by <w>. I have not seen that angled bracket notation before.

From what I have worked on it,
S is a subspace, so $v\in S$. If S = {0}, it is a point. Otherwise there exists a u that is an element of S, where u does not equal the zero vector.
Then the $span\{u\}=\{\lambda u | \lambda \in k\}$, then S is a line. Otherwise there exists a $y\in S$, and $u\in span\{u\}$.
I want to claim u, v are non-collinear, but am not sure how to prove that.

Where I have really gotten stuck though is trying to say:
span{u,v} is a plane and $sp\{u,v\} \subset S \subset k^2$. I claim $sp\{u,v\} = k^2$ so $S = k^2$ and is a plane.

I am supposing u,v are non-collinear. Say u= (a,b) and b=(c,d). If (a,b) and (c,d) each does not equal the zero vector, and $(a,b)\ne \lambda (c,d) \forall \lambda$.
Then let $(\alpha, \beta) \in k^2$. I need to show that there exists $x,y \in k^2$ such that $x(a,b) + y(c,d) = (\alpha, \beta)$.
This is where I really just have no clue. After speaking with my professor, he said it would be a good amount of computation and a 'case analysis', both of which don't help me know what I need to do. Any and all help on this problem would be much appreciated!!

(Sorry for any math typos, first time using LaTex!)

4. Hello
$\mathbb{K}\times \left \{ 0 \right \}$ is a subspace of $\mathbb{K}^2$
which can be written as $\mathbb{K}\times \left \{ 0 \right \}=\left \{ (x,0)\mid x\in \mathbb{K} \right \}=\left \{ x(1,0)\mid x\in \mathbb{K} \right \}$
therefore $\mathbb{K}\times \left \{ 0 \right \}=\textbf{Span}(1,0)$ which means $\mathbb{K}\times \left \{ 0 \right \}$ is a line.
now suppose $\mathbb{E}\subset \mathbb{K}^2$
If $\mathbb{E}$ has a unique element then it must be the zero vector since it's a subspace of $\mathbb{K}^2$.
$\mathbb{E}=\left \{ \vec{0} \right \}$.

5. Originally Posted by Raoh
Hello
$\mathbb{K}\times \left \{ 0 \right \}$ is a subspace of $\mathbb{K}^2$
which can be written as $\mathbb{K}\times \left \{ 0 \right \}=\left \{ (x,0)\mid x\in \mathbb{K} \right \}=\left \{ x(1,0)\mid x\in \mathbb{K} \right \}$
therefore $\mathbb{K}\times \left \{ 0 \right \}=\textbf{Span}(1,0)$ which means $\mathbb{K}\times \left \{ 0 \right \}$ is a line.
now suppose $\mathbb{E}\subset \mathbb{K}^2$
If $\mathbb{E}$ has a unique element then it must be the zero vector since it's a subspace of $\mathbb{K}^2$.
$\mathbb{E}=\left \{ \vec{0} \right \}$.
Thanks. Do you have any insight though on proving how it could be a plane also, and only those three options?

6. Well, how is a plane defined? If your prof. does not expect you to say that a two-dimensional vector space is a plane, what definition of a plane are you expected to use? It would help a lot to know this if we are to answer your question.

7. Hello
$\mathbb{K}^2$ is a subspace of itself.
and we can write $\mathbb{K}^2$ as,
$\mathbb{K}^2=\left \{ (x,y)\mid x,y\in \mathbb{K} \right \}=\left \{ x(1,0)+y(0,1)\mid x,y\in \mathbb{K} \right \}=\texttt{Span}((0,1),(1,0))
$
.
since $\texttt{dim}_{\mathbb{K}}\left (\texttt{Span}((0,1),(1,0)) \right )=2$
$\mathbb{K}^2$ is a plane.