Linear Algebra II

Quote:

Originally Posted by taypez

Need some help on this problem:

Prove that if V is finite dimensional and U1....Um are subspaces of V, then

dim(U1+...+Um) is less than or equal to dim U1+...+Um.

Thanks

Again, I will assume that $U_1+...+U_m$ means the direct sum of vector spaces, as in the Cartesian product.
For simplicity sake, I will work with only two vector spaces. $U_1,U_2\leq V$ . And the more general case will follow easily and similarly.
I did not confirm this because of laziness, but I believe that if $B_1=\{u_1,u_2,...,u_n\}$ is a basis for $U_1$ , and $B_2=\{v_1,v_2,...,v_m\}$ is a basis for $V_2$ . Then, $B_1+B_2=\{(b_1,b_2)|b_1\in B_1\mbox{ and }b_2\in B_2\}$ is not necessarily a basis for $V_1+V_2$ but it is a spanning set of vectors. And hence, since the size of the basis is less than the size of the spanning set we have,
$\dim (V_1+V_2)\leq \dim V_1+\dim V_2=n+m$
Because $n+m$ is the size of spanning set and $\dim (V_1+V_2)$ is the size (cardinality) of the dimension (basis) of those spanning vectors.