What is the smallest possible dimension of $V\cap W$?

Question

Let $V$ and $W$ be subspaces of the vector space $\Bbb{R}^9$ over the field $\Bbb{R}$ of real numbers with $\dim (V) = 5$ and $\dim (W) = 6$.

Then what is the smallest possible dimension of $V\cap W$?

My answer : $\max[\dim(W_1),\dim(W_2)] \le \dim(W_1 ∩ W_2) \le \min[ \dim(W_1) +\dim (W_2), \dim (V)]$

$6\le \dim(W_1\cap W_2) \le 9$

so I got the smallest possible dimension of $V ∩ W$ = $6$.

is its True/false ?

Answer 1

Let $V \subseteq W$, and hence $ V \cap W=V$ and hence I have disproved your claim.

We actually have $$\dim(V \cap W) \le \min(\dim(V), \dim(W)).$$

Consider a basis for $V \cap W$ of $k$ elements, we can then extend to a basis of $V$ using another $(5-k)$ elements and extend it to a basis of $W$ using another $(6-k)$ elements. Not

Hence we will need $$k+(5-k)+(6-k) \le 9$$

$$11-k \le 9$$

$$k \ge 2$$

$$\dim(V \cap W) \ge 2.$$

You might like to explicitly constructed example of $V$ and $W$ to show that it can be attained.

Answer 2

Note that $V\cap W\subset V$, $V\cap W\subset W$ and then $\dim(V\cap W)\leq \min\{\dim (V),\dim (W)\}=5$. Thus, $\dim(V\cap W)$ can not be $6$.

Note that $$9=\dim(\Bbb{R}^9)\geq \dim(V+W)=\dim(V)+\dim(W)-\dim(V\cap W)=11-\dim(V\cap W)$$

Then, $\dim(V\cap W)\geq 2$.