# Bounds on dimension of intersection of three subspaces

Let three distinct subspaces of $\mathbb{R}^{10}$ be $W_1,W_2,W_3$ have dimensions $9$ each. Then, what are the upper and lower bounds on the dimension of $W=\displaystyle\cap_{i=1}^3W_i$? Can we say that $dim (W)=7$? I know that the standard formula for sets does not apply here because the subspaces need not be independent. Any ideas. Thanks beforehand.

Note that if $W_i \neq W_j$ then $(W_i + W_j) = W$. Now use the fact that dim $(U+V) =$ dim $U$ + dim $V$ - dim $U \cap V$ to find dim $W_i \cap W_j$. Finally if $W_i \cap W_j \not \subset W_k$ for $k \neq i,j$ then $W = W_k + W_i \cap W_j$, so apply the dimension formula again.

• how do we know that $W_i\cap W_j$ not a subspace of $W_k$? – vidyarthi Jun 13 '17 at 9:16
• If it is, then $W_1 \cap W_2 \cap W_3 = W_i \cap W_j$, which you've already found. – David Towers Jun 13 '17 at 9:21
• So $W_1 = W_2 = W_3$ gives that the intersection has dimension $9$; $W_i \neq W_j$, but $W_i \cap W_j \subset W_k$ gives that it is 8; and $W_i \neq W_j$, but $W_i \cap W_j \not \subset W_k$ gives that it is 7. – David Towers Jun 13 '17 at 9:24
• how is $W_i+W_j=W$ if $W_i\neq W_j$? – vidyarthi Jun 13 '17 at 9:29
• If $W_i \neq W_j$, then dim $(W_i + W_j) >$ dim $W_i, W_j = 9$ – David Towers Jun 13 '17 at 9:31

Fred gave you an upper bound for the dimension of $W$, if all spaces are the same, so let's work on a lower bound: In $\mathbb{R}^{10}$, we have the orthogonal complement of every space $U$, that satisfies $\dim(U) + \dim(U^{\perp}) = 10$. Now we can compute that $$W^{\perp} = W_1^{\perp} + W_2^{\perp} + W_3^{\perp},$$ as the orthogonal complement always exchanges intersection and sum. By the dimension formula, we know that $\dim(W_i^{\perp}) = 1$ for all $i$, and hence we get $$\dim(W^{\perp}) \leq 3.$$ This allows us to compute the desired lower bound as $$\dim(W) = 10 - \dim(W^{\perp}) \geq 10-3 = 7.$$ Of course you should show that this bound is tight, so you should give spaces $W_i$ that really achieve this bound. Given the construction above, you only need three linearly independent vectors to generate the one-dimensional spaces $W_i^{\perp}$, so it's not that hard to find the example achieving this bound.

edit: Regarding the question on how to show that the orthogonal complement exchanges sum and intersection: We need to show for all subspaces $U,V$ that $$(U\cap V)^{\perp} = U^{\perp} + V^{\perp}.$$ Using that $(U^{\perp})^{\perp} = U$, we can equivalently show $$U^{\perp} \cap V^{\perp} = (U + V)^{\perp}.$$ They can both be done element-wise, but I personally prefer the second one: Let $x \in U^{\perp} \cap V^{\perp}$. Then $<x,u> = <x,v> = 0$ for all $u \in U$ and all $v \in V$. Thus also $<x,u+v> = 0$. But every element of $U+V$ is of the form $u+v$ for $u \in U$ and $v \in V$ and hence we have shown that $x \in (U + V)^{\perp}$. I'm sure you can now do the other direction on your own.

• how can we prove that dimension of orthogonal complement is additive over subspaces? – vidyarthi Jun 13 '17 at 9:08
• I'll edit the answer. – Dirk Jun 13 '17 at 9:11
• great, so this may even apply for finite dimensional hilbert spaces, isnt it? – vidyarthi Jun 13 '17 at 9:18
• The second equation I showed there should work on a really basic level. For the first equation I used $(U^{\perp})^{\perp} = U$, i.e. that the space in question is non-degenerate. I don't see a proof for the first one without that right now, but that doesn't mean that there is none. So as long as your form is non-degenerate, everything is fine, if you have a radical, you might need to do some work first. – Dirk Jun 13 '17 at 10:02

If you choose a basis for $R^{10}$, each of the spaces $W_i$ is represented by a linear equation with 10 unknowns; their intersection is the solution of the system of these three equations, an by Rouché-Capelli theorem the solution has dimension 10 minus the rank of the matrix of the coefficients of the system, and this rank is less or equal than 3 (the number of equations). Thus your conjecture is correct: $dim\ W \geq 7$.