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?

Thanks in advance.

  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 $U\cap W\neq\{0\}$.
  • $\begingroup$ What does the $#$ symbol represent here? $\endgroup$ – Tegernako Dec 12 '18 at 15:23
  • $\begingroup$ It stands for cardinal. $\endgroup$ – José Carlos Santos Dec 12 '18 at 15:24
  • $\begingroup$ I see, we have not used those terms when we studied, but I'll try to translate the same idea with the dimensions theorem we studied. Thank you very much. $\endgroup$ – Tegernako Dec 12 '18 at 15:33
  • $\begingroup$ For the first question, you can just say $\dim V > \dim U + \dim W \ge \dim(U+W)$. $\endgroup$ – Najib Idrissi Dec 12 '18 at 15:35
  • $\begingroup$ @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. $\endgroup$ – 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 $U \cap W = 0_V$.

  • $\begingroup$ 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? $\endgroup$ – Tegernako Dec 12 '18 at 15:18

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