Question related to dimension Here's the question:-

My solution:-

The formula that my solution starts with, which is the dimension of the addition of two subspaces is already proved in the book. I merely used it in order to prove my result. However, while looking at the solution, I found out that my result is WRONG! There is a counterexample! So can someone tell me where did I go wrong in this solution? I would appreciate any feedback. Thank you.
Btw in the previous chapter, I proved that the sum of two subspaces is associative so I don't think that is the mistake. My guess would be that the Venn Diagrams do not apply to subspaces or the intersection of subspaces is not distributive?
 A: Let $V=\Bbb{R}^2$ and let $U_1=\text{Span}(\{(1,0)\}), U_2=\text{Span}(\{(0,1)\})$ and $U_3=\text{Span}(\{(1,1)\})$. Then
\begin{align}
\text{dim}(U_1+U_2+U_3)&=\text{dim}(\Bbb{R}^2)=2\\
\text{dim}(U_i)&=1 & (\text{ for all } i\in \{1,2,3\})\\
\text{dim}(U_i \cap U_j)&=0 & (\text{ for } i \neq j)\\
\text{dim}(U_1 \cap U_2 \cap U_3)&=0\\
\end{align}
So this does not satisfy the given equality. Hence the result is not true in general.

Note your step
$$\text{dim}((U_1+U_2) \cap U_3)= \text{dim}((U_1 \cap U_3)+(U_2 \cap U_3))$$
is where things go wrong. Because you can see from the example I have given that this is not necessarily true.
A: I think the place you went wrong was when you used:
$$\dim((U_1+U_2)\cap U_3)$$
$$=\dim((U_1\cap U_3)+(U_1\cap U_3))$$
which is not true.
Consider $U_1=\{(x,y) | x=0\}$, $U_2=\{(x,y) | y=0\}$, and $U_3=\{(x,y) | x=y\}$.
Then
$$(U_1+U_2)\cap U_3=\mathbb{R}^2\cap U_3=U_3$$
But
$$(U_1\cap U_3)+(U_1\cap U_3)=\{0\}+\{0\}=\{0\}$$
A: Thanks for all your replies. I found the problem in my solution. It seems the intersection of subspaces is not distributive (which my proof above assumes). Coincidentally, I just got recommended this thread:-
An example of 3 subspaces of $V$ such that $w_1 \cap (w_2+w_3) \neq (w_1 \cap w_2) + (w_1 \cap w_3)$
