Subspaces

**pakman** · May 6th 2008, 07:23 PM

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.

**ThePerfectHacker** · May 6th 2008, 07:33 PM

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

**pakman** · May 6th 2008, 07:35 PM

Originally Posted by ThePerfectHacker

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

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

**ThePerfectHacker** · May 6th 2008, 07:45 PM

Originally Posted by pakman

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

No. Remember the dimension of $\mathbb{R}^3$ over $\mathbb{R}$ is $3$ .
This means that you cannot have a set with more than $3$ 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 $S = \{ (a,0,b)|a,b\in \mathbb{R}\}$ and $T = \{ (a,b,0)|a,b\in \mathbb{R}\}$ .
Do the following steps.
1)Show $S,T$ are sub-spaces.
2)Show $S\cup T$ is not closed, i.e. there are $\bold{x},\bold{y}\in S\cup T$ so that $\bold{x}+\bold{y}\not \in S\cup T$
3)Conclude that $S\cup T$ is not a sub-space.

**pakman** · May 6th 2008, 09:46 PM

Originally Posted by ThePerfectHacker

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

That makes sense now, thanks!

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

**ThePerfectHacker** · May 7th 2008, 09:37 AM

Originally Posted by pakman

Does that work?

Yes.

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