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

  • Thread starter Jamin2112
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  • #1
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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?
 

Answers and Replies

  • #2
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Hi Jamin2112! :smile:

Take a basis of W, can you extend this basis to form a basis of V??
 
  • #3
986
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Hi Jamin2112! :smile:

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)?
 
  • #4
22,089
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No, I'm not. I doubt that a proof by contradiction will be the most efficient route here :frown:
 
  • #5
986
9
Hi Jamin2112! :smile:

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.
 
  • #6
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Can't we just use the fact that every element in W is in V??
 
  • #7
HallsofIvy
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Homework Helper
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Hi Jamin2112! :smile:

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 [itex]R^2[/itex] 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 [itex]R^2[/itex].
 

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