# Subspaces and dimension

subspaces and dimension!!!!

Consider two subspaces V and W of R^n ,where V is contained in W.
Why is dim(V)<= dim(W).....?
"<=" less than or equal to

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the only way you get dim(V) = dim(W) is if V=W, if V is strictly contained in W then, then there must be some vector in W that is not in V, let this be w. Now let (v_1,v_2,...v_r) be a basis of V (now I assumed dim(W)=r). Clearly v_1,v_2,...v_r are in W because it containes V, and because w is not in V it must be independent of v_1,v_2,...v_r, so
(v_1,v_2,...v_r,w) is a linearly independent set in W, so dim(W) is at least r+1 (could be more).

HallsofIvy
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Just because I can't resist "putting my oar in", I'll echo mrandersdk: If V is a subspace of W, then any vector any V is also a vector in W. Any basis for W spans V and so any basis for V cannot be larger than a basis for W. Since the dimension of a space is the size of a basis, ....

By the way, do you see why saying "subspace V is a subset of subspace W" (they are both subspaces of some vector space U) is the same as saying "V is a subspace of W"?