Thread: Subspaces

I need to prove that these statements are true or false.

1. It is possible to find a pair of two-dimensional subspaces S and T of R³ such that S ∩ T = {0}

2. If S and T are subspaces of a vector space V, then S U T is a subspace of V.

3. If S and T are subspace of a vector space V, then S ∩ T is a subspace of V.

I understand that the two closure identities need to be met in order to be a subspace. However the intersection and union properties are throwing me off. Can anyone help me with this? Thank you.

Originally Posted by pakman

1. It is possible to find a pair of two-dimensional subspaces S and T of R³ such that S ∩ T = {0}

Let $\displaystyle \{ \bold{a}_1,\bold{a}_2\}$ be a basis for $\displaystyle S$. And $\displaystyle \{ \bold{b}_1,\bold{b}_2\}$ be a basis for $\displaystyle T$. Note that $\displaystyle \bold{b}_1 \not \in \text{spam}\{ \bold{a}_1,\bold{a}_2\}=S$ because $\displaystyle S\cap T = \{ \bold{0} \}$. Similarly, $\displaystyle \bold{b}_2\not \in \text{spam}\{ \bold{a}_1,\bold{a}_2\}$. This means the set $\displaystyle \{ \bold{a}_1,\bold{a}_2,\bold{b}_1,\bold{b}_2\}$ is a linearly independent set. That is a contradiction. Why?

Originally Posted by ThePerfectHacker

Let $\displaystyle \{ \bold{a}_1,\bold{a}_2\}$ be a basis for $\displaystyle S$. And $\displaystyle \{ \bold{b}_1,\bold{b}_2\}$ be a basis for $\displaystyle T$. Note that $\displaystyle \bold{b}_1 \not \in \text{spam}\{ \bold{a}_1,\bold{a}_2\}=S$ because $\displaystyle S\cap T = \{ \bold{0} \}$. Similarly, $\displaystyle \bold{b}_2\not \in \text{spam}\{ \bold{a}_1,\bold{a}_2\}$. This means the set $\displaystyle \{ \bold{a}_1,\bold{a}_2,\bold{b}_1,\bold{b}_2\}$ is a linearly independent set. That is a contradiction. Why?

Because S and T are a linear combination to get 0?

Originally Posted by pakman

Because S and T are a linear combination to get 0?

No. Remember the dimension of $\displaystyle \mathbb{R}^3$ over $\displaystyle \mathbb{R}$ is $\displaystyle 3$.
This means that you cannot have a set with more than $\displaystyle 3$ elements which are linearly independent.

2. If S and T are subspaces of a vector space V, then S U T is a subspace of V.

Let $\displaystyle S = \{ (a,0,b)|a,b\in \mathbb{R}\}$ and $\displaystyle T = \{ (a,b,0)|a,b\in \mathbb{R}\}$.
Do the following steps.
1)Show $\displaystyle S,T$ are sub-spaces.
2)Show $\displaystyle S\cup T$ is not closed, i.e. there are $\displaystyle \bold{x},\bold{y}\in S\cup T$ so that $\displaystyle \bold{x}+\bold{y}\not \in S\cup T$
3)Conclude that $\displaystyle S\cup T$ is not a sub-space.

Originally Posted by ThePerfectHacker

No. Remember the dimension of $\displaystyle \mathbb{R}^3$ over $\displaystyle \mathbb{R}$ is $\displaystyle 3$.
This means that you cannot have a set with more than $\displaystyle 3$ elements which are linearly independent.

That makes sense now, thanks!

Let $\displaystyle S = \{ (a,0,b)|a,b\in \mathbb{R}\}$ and $\displaystyle T = \{ (a,b,0)|a,b\in \mathbb{R}\}$.
Do the following steps.
1)Show $\displaystyle S,T$ are sub-spaces.
2)Show $\displaystyle S\cup T$ is not closed, i.e. there are $\displaystyle \bold{x},\bold{y}\in S\cup T$ so that $\displaystyle \bold{x}+\bold{y}\not \in S\cup T$
3)Conclude that $\displaystyle S\cup T$ is not a sub-space.

s = (1,0,1) and u = (1,1,0) then sUv = (2,2,1) which does not belong to either s or u. Does that work?

Originally Posted by pakman

Does that work?

Yes.