Let $U, W$ be subspaces of finite dimensional Vector Space $V$ over $K$ with $\dim U + \dim W > \dim V$. Show that $U \cap W \neq$ {$0$}.

Question

Let $U, W$ be subspaces of finite dimensional Vector Space $V$ over $K$ with $\dim U + \dim W > \dim V$. Show that $U \cap W \neq$ {$0$}.

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The first thing I thought about is this formula $$\dim(U+W) = \dim U + \dim W - \dim (U \cap W)$$

We know $\dim(U+W) \leq \dim V$ because $U,W$ are subspaces of $V$, and the dimension of a subspace is always less than or equal to the dimension of the space they belong to.

Using those two formulas, we have $$\dim V \geq \dim(U+W) = \dim U + \dim W - \dim (U \cap W)$$

If we rearrange, we have $$\dim (U \cap W) \geq \dim U + \dim W - \dim (V)$$

We are given that
$$\dim U + \dim W > \dim V$$ so we know $$\dim U + \dim W - \dim (V) > 0$$ so we have $$\dim (U \cap W) \geq \dim U + \dim W - \dim (V) > 0$$ which implies $U \cap W \neq$ {$0$}.

Is this correct or did I do something wrong or forgot something ?

Answer 1

Your proof is correct. You can also use proof by contradiction.

Suppose, for a contradiction, that $U\cap W=\{0\}$. Then $\dim(U\cap W)=0$. Since $\dim(U+W)=\dim U+\dim W-\dim(U\cap W)$, we have $\dim(U+W)=\dim U+\dim W$. We know that $\dim V\geq\dim(U+W)$, because $U+W$ is a subspace of $V$. It follows that $\dim V\geq\dim U+\dim W$, which contradicts our assumption that $\dim V<\dim U+\dim W$. This completes the proof.