dimension

Kat-M · #1 $Old$ January 11th, 2009, 08:55 PM

Let V be a finite dimensional vector space. Show that if W1,....,Wn are subspaces of V such that none of these subspaces are qeual to V, then Union of all these subspaces does not equal V.

i am thinking induction on dim V might help but do not really have idea how to go bout it. so please help.

ThePerfectHacker · #2 $Old$ January 11th, 2009, 11:12 PM

Quote:

Originally Posted by Kat-M $View Post$

Let V be a finite dimensional vector space. Show that if W1,....,Wn are subspaces of V such that none of these subspaces are qeual to V, then Union of all these subspaces does not equal V.

The union of two subpaces is a subspace if and only one is contained in another.
Thus if $W_1\cup W_2$ is a subspace thus $W_1\subseteq W_2$ and $W_2\subseteq W_1$ .
And so $\dim (W_1\cap W_2) = \max \{ \dim (W_1),\dim (W_2) \} < \dim (V)$ .

HallsofIvy · #3 $Old$ January 12th, 2009, 07:42 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

The union of two subpaces is a subspace if and only one is contained in another.

The reason that is so: Suppose u is in subspace $U_1$ but NOT subspace $U_2$ and v is in subspace $U_2$ but not in $U_1$ . The u+ v cannot be in $U_1\cup U_2$ . If it were, then it would have to be in either $U_1$ or $U_2$ (or both). If u+ v were in $U_1$ , then, because $U_1$ is closed under addition and scalar multiplication u+ v+ (-u)= v would be in $U_1$ , a cotradiction. If u+ v were in $U_2$ , simlarly u+ v+ (-v)= u would be in $U_2$ , again a contradiction.