If W is a subset of V, then dim(W) ≤ dim(V)

[Jamin2112]

Homework Statement

I need to prove this:

W, V are linear subspaces
W is a subset of V

-----> dimension(W) ≤ dimension(V)

Homework Equations

dimension(X): # of linearly independent vectors in any basis of X

The Attempt at a Solution

I'm trying to think this through, but getting stalled.

Hmmmm...

Suppose dim(W) > dim(V). Given any basis of W and any basis of V, there will be some vector w* such that w* is contained in the basis of W but not in the basis of V.


... somehow I'm supposed to deduce a contradiction (if this is even the most efficient way to the conclusion).

Help?

 

[micromass]

Hi Jamin2112!

Take a basis of W, can you extend this basis to form a basis of V??

 

[Jamin2112]

Hi Jamin2112!

Take a basis of W, can you extend this basis to form a basis of V??

Are you talking about my supposition where dim(W)>dim(V)?

 

[micromass]

No, I'm not. I doubt that a proof by contradiction will be the most efficient route here

 

[Jamin2112]

Hi Jamin2112!

Take a basis of W, can you extend this basis to form a basis of V??

If dim(W) = dim(V), yes;
if dim(W) < dim(V), no.

 

[ManiFresh]

Can't we just use the fact that every element in W is in V??

 

[HallsofIvy]

Hi Jamin2112!

Take a basis of W, can you extend this basis to form a basis of V??

If dim(W) = dim(V), yes;
if dim(W) < dim(V), no.

Why not? The x-axis is a subset of $R^2$ and a vector space of dimension 1. It has {<1, 0>} as basis. Adding <0, 1> to that set gives {<1, 0>, <0, 1>}, extending the first basis to a basis of $R^2$.