# Possible dimensions for a sum of two vector spaces

Let $U, W \subseteq \mathbb{R}^6$ vector subspaces from dimension $4$. Show that $U \cap W$ holds at least $2$ linearly independent vectors and at most $4$ linearly independent vectors.

Using the dimensions theorem I showed by assuming the opposite that $0 \leq dim(U+W) \leq 3$ or $7 \leq dim(U+W) \leq 8$

But I'm not sure on why this is a contradiction. I understand that if $7 \leq dim(U+W) \leq 8$ than there's no possible way a subspace of $\mathbb{R}^6$ would have a base that has $\geq 7$ vectors. I'm not sure why that base can't have $3$ vectors or less.

• Should "linearly dependent" actually be "linearly independent"? Because given any non-zero vector $v$ in $U\cap W$, I can easily give you infinitely many linearly dependent vectors: $2v$, $3v$, $4v$, $5v$, $6v$, … – celtschk Jun 1 '13 at 13:18
• @celtschk ,yes I'm sorry for the confusion. – Georgey Jun 1 '13 at 13:22
• Well, the "at most $4$" part is easy by noticing that any set of linear independent vectors in $U\cap W$ is also a set of linearly independent vectors in $U$. – celtschk Jun 1 '13 at 13:32
• and also for $W$, right? But is the other part of the proof ok? – Georgey Jun 1 '13 at 13:39
• Yes, also for $W$ (but since $\dim W=\dim U$, any one of them suffices in this case). And yes, as far as I can tell, the rest of your proof should be OK. – celtschk Jun 1 '13 at 13:47

## 2 Answers

You should know the basic result about this, namely Grassmann's formula $$\dim(U+W)=\dim U+\dim W-\dim(U\cap W).$$ Since $U$ and $W$ are four dimensional subspaces of $\mathbb{R}^6$, you surely have $$4\le\dim(U+W)\le 6$$ Can you go on from here?

• Why would $4 \leq dim(U+W) \leq 6$? – Georgey Jun 1 '13 at 14:10
• Because $4=\dim U$ and $6=\dim \Bbb R^6$ – xavierm02 Jun 1 '13 at 14:16
• Because $U\subseteq U+W\subseteq\mathbb R^6$. – Andreas Blass Jun 1 '13 at 14:16
• @Georgey As Andreas said, since $U\subseteq U+W$, you have $\dim U\le \dim (U+W)$. The dimension of $U+W$ cannot exceed $6$, because it's a subspace of $\mathbb{R}^6$. – egreg Jun 1 '13 at 14:18

This solution is based on elementary theory of linear systems of equations.

Take $\mathbf x \in U \cap W$. Then $\mathbf x=\sum a_i\mathbf u_i=\sum b_j\mathbf w_j$, where $\mathbf u_i$ and $\mathbf w_j, i=1,...,4, j=1,...,4$, are vectors that form a basis of $U$ and $W$. We look at the solutions of the homogeneous linear system $$\sum a_i\mathbf u_i-\sum b_j\mathbf w_j=\mathbf 0.$$ We have 6 equations, as $\mathbf x\in R^6$, and 8 unknowns (4 values for the $a_i$ and 4 values for the $b_i$). Notice that the matrix $A$ of coefficients of the homogeneous system has dimension 6 x 8. Hence, the rank of $A$ is $4\leq r(A)\leq 6$, as the rank is 4 when all the $w_j$s linearly depend on the $u_i$ and it cannot exceed the number of rows which is 6. By Rouché-Capelli theorem, the system always have solutions that depend on $k=8-r(A)$ free parameters. Clearly, we have: $$2\leq k\leq 4,$$
and this means that any solution $\mathbf x$ can be written as a linear combination of two to four linearly independent vectors.