# Claims regarding dimensions of vector space and subspaces

I have to check whether the following claims are true or false:

Let $$V$$ be a vector space of finite dimension, and let $$U, W$$ be subspaces of $$V$$.

1) If $$\dim V > \dim U + \dim W$$, then $$V \neq U + W$$

2) If $$\dim V > \dim U + \dim W$$, then $$U\cap W = 0_v$$

I have been struggling with this material since we first started it.

From what I can gather, the first one is correct, but this is only based on my intuition.

For the second claim, what I think is:

We know that $$\dim (U +W) = \dim U + \dim W - \dim (U \cap W)$$.

Combining this with the information we have from the claim, we can assume:

$$\dim (U \cap W)= 0$$.

Therefore $$U \cap W = {0v}$$.

Is this a valid way of proofing?

1. Suppose that $$U\cap W=\{0\}$$. Then, if $$B_U$$ is a basis of $$U$$ and $$B_W$$ is a basis of $$W$$, then $$B_U\cup B_W$$ is linearly independent and therefore\begin{align}\dim V&\geqslant\#(B_U\cup B_W)\\&=\#B_U+\#B_W\\&=\dim U+\dim W.\end{align}And if $$U\cap W\neq\{0\}$$. Let $$B_{U\cap W}$$ be a basis of $$U\cap W$$, and extend it to a basis $$B_U$$ of $$U$$ and to a basis $$B_W$$ of $$W$$. Then\begin{align}\dim V&\geqslant\#(B_U\cup B_W)\\&=\#B_U+\#B_W-\#B_{U\cap W}\\&>\dim U+\dim W.\end{align}
2. The statement is false. Let $$V=\mathbb{R}^3$$, let $$v\in V\setminus\{0\}$$ and let $$U=W=\mathbb{R}v$$. Then$$3=\dim V>1+1=\dim U+\dim W,$$but $$U\cap W\neq\{0\}$$.
• What does the $#$ symbol represent here? – Tegernako Dec 12 '18 at 15:23
• For the first question, you can just say $\dim V > \dim U + \dim W \ge \dim(U+W)$. – Najib Idrissi Dec 12 '18 at 15:35
• @NajibIdrissi Of course, but I assumed that the OP was not aware of the fact that $\dim(U+W)\leqslant\dim U+\dim W$. With that knowledge, the problem becomes trivial. – José Carlos Santos Dec 12 '18 at 15:37
There is a contradiction example for the second statement. suppose $$dim(V) = 10$$ and $$(U \cup W) \subset V$$ and $$U$$ and $$V$$ have intersection with each other such that $$U = W$$ and $$dim(U) = dim(W) = 1$$, but not $$U \cap W = 0_V$$.
• I see, thanks. Suppse this was the other way around, I mean if $dimV < dimU + dimW$, therefore $U \cap W \neq {0}$ was necessarily true, from definitions and what we have, right? – Tegernako Dec 12 '18 at 15:18