Let $ V $ be a linear space over a field $ F $ and let $ W, W_1, W_2 $ be subspaces of $ V$. Assume that $ V = W \oplus W_1 = W \oplus W_2. $ Prove that $$\dim(W_1 \cap W_2) \ge \dim V - 2 \dim W.$$
Could you, please, verify the following proof? $$\dim W_1 + \dim W = \dim V \\ \dim W_2 + \dim W = \dim V$$ \begin{align} \dim(W_1 + W_2) &= \dim W_1 + \dim W_2 - \dim (W_1 \cap W_2) \\ &= \dim V - \dim W + \dim V - \dim W - \dim(W_1 \cap W_2) \\ &= 2 \dim V - 2 \dim W - \dim(W_1 \cap W_2) \end{align} $$\dim(W_1 + W_2) + \dim(W_1 \cap W_2) = 2 \dim V - 2 \dim W$$ Since $W_1 + W_2$ is a subspace of $V$, we have $\dim(W_1 + W_2) \le \dim V$ and, therefore, we get: $$\dim(W_1 \cap W_2) \ge \dim V - 2 \dim W.$$