Help on some Linear Algebra - subspaces

• Feb 8th 2011, 03:04 PM
Lord Voldemort
Help on some Linear Algebra - subspaces
Mainly I just want verification if this is correct or otherwise.

Consider two subspaces V and W of R_n.
Is V U W necessarily a subspace of R? Is the intersect of V and W a subspace?

Yes to both?
If a vector x is an element of V intersect W, then it is an element of V, then therefore all linear combinations are also in V. It is also in W, and therefore all linear combinations are in W as well. Thus all linear combinations are in V intersect W. The other one would use similar logic.

ALSO

Consider vectors v1, v2... v_m in |Rn. Is the span of the vectors necessarily a subspace of Rn?

I think this is true but how would I justify it?
Say that span is all linear combinations, and then c can be 0 so that fulfills one requirement.
Then say that all scalar multiples of any vectors with real numbers will result in vectors of real numbers?
• Feb 8th 2011, 03:15 PM
dwsmith
Quote:

Originally Posted by Lord Voldemort
Mainly I just want verification if this is correct or otherwise.

Consider two subspaces V and W of R_n.
Is V U W necessarily a subspace of R? Is the intersect of V and W a subspace?

Yes to both?
If a vector x is an element of V intersect W, then it is an element of V, then therefore all linear combinations are also in V. It is also in W, and therefore all linear combinations are in W as well. Thus all linear combinations are in V intersect W. The other one would use similar logic.

ALSO

Consider vectors v1, v2... v_m in |Rn. Is the span of the vectors necessarily a subspace of Rn?

I think this is true but how would I justify it?
Say that span is all linear combinations, and then c can be 0 so that fulfills one requirement.
Then say that all scalar multiples of any vectors with real numbers will result in vectors of real numbers?

If the vectors in V and W are subspaces of R^n, the union of of course will be subspace.

The intersection of V and W could be empty. Is an empty set a subspace? I am not too sure of that.
• Feb 8th 2011, 03:36 PM
Lord Voldemort
Thanks!
V intersect W is not necessarily a subspace, since the empty set does not contain 0.
• Feb 8th 2011, 03:37 PM
dwsmith
Quote:

Originally Posted by Lord Voldemort
Thanks!
V intersect W is not necessarily a subspace, since the empty set does not contain 0.

I was to lazy to look up anything so goodjob.
• Feb 8th 2011, 06:51 PM
Ackbeet
If V and W are both subspaces, they both contain the zero vector, and hence the intersection is nonempty.
• Feb 8th 2011, 08:21 PM
TheEmptySet
Quote:

Originally Posted by dwsmith
If the vectors in V and W are subspaces of R^n, the union of of course will be subspace.

The intersection of V and W could be empty. Is an empty set a subspace? I am not too sure of that.

I have to disagee. The union of two subspaces is, NOT in general, a subspace. Here is a very simple counter example.

Lets consider $\displaystyle \mathbb{R}^2$

Now consider the subspace spanned by $\displaystyle V=\vec{i}$ (the x axis)

and the subspace spanned by $\displaystyle W=\vec{j}$ (the y axis)

Now the Union of these two subspaces is both coordinate axes. This is not a vector space at $\displaystyle \displaystyle \vec{i}+\vec{j} \notin V \cup U$

What is true is if you take the sum $\displaystyle V+W=\{v+w,v\in V w \in W\}$ this is a subspace.
• Feb 8th 2011, 08:43 PM
Drexel28
Quote:

Originally Posted by TheEmptySet
I have to disagee. The union of two subspaces is, NOT in general, a subspace. Here is a very simple counter example.

Lets consider $\displaystyle \mathbb{R}^2$

Now consider the subspace spanned by $\displaystyle V=\vec{i}$ (the x axis)

and the subspace spanned by $\displaystyle W=\vec{j}$ (the y axis)

Now the Union of these two subspaces is both coordinate axes. This is not a vector space at $\displaystyle \displaystyle \vec{i}+\vec{j} \notin V \cup U$

What is true is if you take the sum $\displaystyle V+W=\{v+w,v\in V w \in W\}$ this is a subspace.

In fact, if $\displaystyle U,W\leqslant V$ then $\displaystyle U\cup W\leqslant V$ if and only if $\displaystyle U\subseteq W$ or $\displaystyle W\subseteq U$. Indeed, suppose neither containment held, then we may choose $\displaystyle u\in U-W$ and $\displaystyle w\in W-U$. A quick check then shows that $\displaystyle u+w\notin U$ since otherwise $\displaystyle (u+w)-u=w\in U$ contrary to construction. Similarly, $\displaystyle u+w\notin W$. Thus, $\displaystyle u+w\notin U\cup W$ etc. In fact, this is really just a theorem about the underlying group structure of the vector spaces.
• Feb 8th 2011, 08:51 PM
Drexel28
Quote:

Originally Posted by TheEmptySet

What is true is if you take the sum $\displaystyle V+W=\{v+w,v\in V w \in W\}$ this is a subspace.

It's also interesting to note that this is not strange since, if $\displaystyle \mathcal{V}$ is the ambient space, $\displaystyle \displaystyle V+W=\bigcap_{\substack{U\leqslant \mathcal{V}\\ V\cup W\subseteq U}}U$