Intersecting Vector Spaces

Let $U,V,W$ be distinct lines through the origin in ${\mathbb R}^2.$ Then $U+V+W = {\mathbb R}^2$ but all the intersections are zero-dimensional, so the left side is $2$ but the right side is $3.$ So the statement is false.

It is true that $\begin{aligned} \dim(U+V+W) &= \dim(U+V) + \dim W - \dim((U+V) \cap W) \\ &= \dim U + \dim V - \dim(U\cap V) + \dim W - \dim((U+V) \cap W), \end{aligned}$ so the statement fails precisely when $\dim((U+V) \cap W) \ne \dim(U\cap W) + \dim(V\cap W) - \dim(U \cap V \cap W).$ One way to make this fail is when $U+V$ intersects $W$ nontrivially but $U$ and $V$ do not, as in the counterexample above.

First up, the 1st formula is not in general valid. It only holds for finite dimensional vector-spaces. Otherwise the formula is:

$\dim(U+V) + \dim(U\cap V) = \dim U+\dim V.$

I would also like to comment on how to approach this problem without immediately constructing a counter-example (synthetic thinking), but by taking analytical thinking as far as it goes, until the problem remains open, and synthetic thinking becomes necessary (and is, thanks to the analytical steps, better informed). The problem with the other approaches is, they just jump straight to counter-examples. That’s fine for humanity to know whether or not something is true, but it fails to explain how to get to these otherwise magically appearing solutions.

Step I. Working with finite dimensional spaces, utilising the given valid identity, one has

$\begin{array}{c}[c]{rcl} \dim U+V+W = \dim (U+V)+W &= &\dim(U+V) + \dim W - \dim ((U+V)\cap W)\\ &= &\begin{array}{c}[t]{l} \dim U + \dim V - \dim (U\cap V)\\ +\dim W - \dim ((U+V)\cap W)\\ \end{array}\\ &= &\dim U + \dim V + \dim W - \dim (U\cap V) - \dim ((U+V)\cap W).\\ \end{array}$

Step 2. One could go with gut feeling at this stage … and assume that by symmetry the 2nd formula in the OP must hold. But intuition and gut feeling have no decisive weight in mathematics. We would still need back this up by a proof. Looking at what has been analytically derived so far, it is clear what the crucial question is, it is namely: Does

$\dim ((U+V)\cap W) = \dim (U\cap W) + \dim (V\cap W) - \dim(U\cap V\cap W)$

hold? The answer to the OP is decided by the validity or invalidity this one formula.

Step 3 (cross-over to synthetic thinking based on analytical thinking). Now lets us reflect a little, if (big if) it were the case, that (†) $(U+V)\cap W = (U\cap W) + (V\cap W)$ , then this last identity would hold, since in particular $(U\cap W)\cap (V\cap W) = U\cap V\cap W$ . That is, (†) is an obvious sufficient condition . Now clearly,

$(U\cap W) + (V\cap W) \subseteq (U+V)\cap W$

Supposing, now we could show, that condition (†) did not hold, then $(U\cap W) + (V\cap W)\subset (U+V)\cap W$ strictly. Since all spaces here are finite-dimensional, this would imply

$\begin{array}{c}[c]{rcl} \dim ((U+V)\cap W) &> &\dim ((U\cap W) + (V\cap W))\\ &= &\dim (U\cap W) + \dim (V\cap W) - \dim((U\cap W)\cap (U\cap W)\\ &= &\dim (U\cap W) + \dim (V\cap W) - \dim(U\cap V\cap W)\\ \end{array}$

Hence, condition (†) is not only sufficient, it is necessary for the 2nd formula in the OP. We have thus reduced the problem to solving whether or not

$(U+V) \cap W \subseteq (U\cap W) + (V\cap W)$

in general holds for all finite-dimensional vector spaces. (The other inclusion is true, and together they are equivalent to (†).)

Step 4 (analytical step). This problem is furthermore clearly equivalent to determining whether or not

a basis $B$ , for $(U+V)\cap W$ , such that $B=B_{U}\cup B_{V}$ with $B_{U}\subseteq U$ and $B_{V}\subseteq V$

exists.

Step 5 (synthetic step). Roughly speaking, the latter holding true means $W$ polarises $U+V$ into $U$ and $V$ . But this is clearly not always the case: $W$ may irreversibly entangle the elements of $U$ with those of $V$ , and this polarisation is in general irreversible. This leads us to the construction of the simplest possible counter-example: setting $U=\langle e_{0}\rangle$ , $V=\langle e_{1}\rangle$ and $W=\langle e_{0}+e_{1}\rangle$ in some arbitrary $d>2$ -dimensional over-space with basis \(\{e_{0},e_{1},\ldots\}), we have

a basis for \((U+V) \cap W\) is necessarily of the form $B=\{c\cdot(e_{0}+e_{1})\}$ for some scalar $c$ . But clearly $c\cdot(e_{0}+e_{1})\notin U$ and $c\cdot(e_{0}+e_{1})\notin V$ .

Thus, in the reduced problem the answer is No . Hence the answer to the original problem is No .

Consider the following subspaces of $\mathbb{R}^3$ : $U \; = \; \langle e_1,e_2 \rangle \hspace{1cm} V \; = \; \langle e_1,e_3 \rangle \hspace{1cm} W \; = \; \langle e_1,e_2+e_3 \rangle$ where $e_1,e_2,e_3$ forms the canonical basis for $\mathbb{R}^3$ . Then $U + V + W = \mathbb{R}^3$ , while $U \cap V \; = \; U \cap W \; = \; V \cap W \; = \; U \cap V \cap W \; = \; \langle e_1 \rangle$ and so $U,V,W$ all have dimension $2$ , while $U \cap V, U \cap W, V \cap W, U \cap V \cap W$ all have dimension $1$ . Thus in this case, $\begin{aligned} \mathrm{dim}\,(U + V + W) & = \; 3 \\ \mathrm{dim}\,U + \mathrm{dim}\,V + \mathrm{dim}\,W - \mathrm{dim}\,U \cap V - \mathrm{dim}\,U \cap W - \mathrm{dim}\,V \cap W + \mathrm{dim}\, U \cap V \cap W & = \; 2 + 2 + 2 - 1 - 1 - 1 + 1 \; = \; 4 \end{aligned}$ so the stated identity is, in general, false.