1. ## [SOLVED] Dimension

$\displaystyle U_1 \cap U_2 = \{0\}$, $\displaystyle U_1 \subset V$, $\displaystyle U_2 \subset V$

Show that $\displaystyle dim(U_1)+dim(U_2) \leq dim(V)$

2. Originally Posted by Spec
$\displaystyle U_1 \cap U_2 = \{0\}$, $\displaystyle U_1 \subset V$, $\displaystyle U_2 \subset V$

Show that $\displaystyle dim(U_1)+dim(U_2) \leq dim(V)$
Let $\displaystyle W$ be the sum of vector subspaces $\displaystyle U_1$ and $\displaystyle U_2$.
Then $\displaystyle W\subseteq V$ and so $\displaystyle \text{dim} W \leq \text{dim} V$.
However, $\displaystyle \text{dim} U_1 + \text{dim} U_2 = \text{dim} W$.

3. Originally Posted by ThePerfectHacker
Let $\displaystyle W$ be the sum of vector subspaces $\displaystyle U_1$ and $\displaystyle U_2$.
Then $\displaystyle W\subseteq V$ and so $\displaystyle \text{dim} W \leq \text{dim} V$.
However, $\displaystyle \text{dim} U_1 + \text{dim} U_2 = \text{dim} W$.
(1) $\displaystyle dim(U_1\oplus U_2) = dim U_1+dim U_2$
(2) $\displaystyle U_1\oplus U_2 \subseteq V$

$\displaystyle (1), (2) \implies \text{dim} U_1 + \text{dim} U_2 \leq \text{dim} V$

Is that proof enough? I can prove (1), but not (2).

4. Originally Posted by Spec
Is that proof enough? I can prove (1), but not (2).
Remember what the the sum of subspaces are. It is the smallest subspace that contains both $\displaystyle U_1$ and $\displaystyle U_2$.
However, $\displaystyle V$ already contains $\displaystyle U_1,U_2$, so $\displaystyle W\subseteq V$.