# Thread: [SOLVED] Dimension

1. ## [SOLVED] Dimension

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

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

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

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

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

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

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 $U_1$ and $U_2$.
However, $V$ already contains $U_1,U_2$, so $W\subseteq V$.