Proof Question involving subspaces in Linear Algebra

Question

Suppose that $U$ and $W$ are subspaces of a vector space $V$ . Prove that if $\dim U + \dim W > \dim V$, then $U \cap W \neq \{0\}.$

How can I go about proving this? I know the identity that:

• $\dim( U + W) + \dim (U ∩ W) = \dim(U) + \dim(W)$ , since $U$ and $W$ are both subspaces of $V$ but I'm not too sure how to approach this. Any help would be appreciated. Thanks.

Answer 1

You know the Grassmann formula, just supplement it with the last inequality below: $$ \dim U+\dim W-\dim(U\cap W)=\dim(U+W)\le\dim V $$ Then you can immediately deduce that $$ \dim(U\cap W)\ge\dim U+\dim W-\dim V $$ and, in your case, $$ \dim(U\cap W)>0 $$

Answer 2

By hypothesis $\dim(U+W)+\dim(U\cap W)=\dim U+\dim W>\dim V$. If $U\cap W=\emptyset$ then $\dim(U\cap W)=0$ then $\dim(U+W)>\dim V$, absurd because $U+W$ is a subspace of $V$ then $\dim(U+W)\leq \dim V$. Therefore, $U\cap W\neq\emptyset$.

Answer 3

One could go about it by proving the contrapositive.

Prove that if $(U \cap W) = \lbrace \mathbf{0} \rbrace$, then $\text{dim}(U) + \text{dim}(W) \leq \text{dim}(V)$.

Proof suggestion:

Let us assume $(U \cap W) = \lbrace \mathbf{0} \rbrace$. We know that if $(U \cap W) = \lbrace \mathbf{0} \rbrace$, then $\text{dim}(U \cap W) = 0$. We also know that $\text{dim}(U + W) \leq \text{dim}(V)$. If we combine these two, we get that $$ \text{dim}(U + W) + \text{dim}(U \cap W) \leq \text{dim}(V). $$ Since we also know that $\text{dim}(U + W) + \text{dim}(U \cap W) = \text{dim}(U) + \text{dim}(W)$, this is the same as $$ \text{dim}(U) + \text{dim}(W) \leq \text{dim}(V). $$

Let us now assume that $\text{dim}(U) + \text{dim}(W) > \text{dim}(V)$. If so, then it can't be the case that $(U \cap W) = \lbrace \mathbf{0} \rbrace$, because that would imply $\text{dim}(U) + \text{dim}(W) \leq \text{dim}(V)$. We can then conclude that $(U \cap W) \neq \lbrace \mathbf{0} \rbrace$.