Provide an example of three subspaces $\mathcal{U}, \mathcal{V}$ and $\mathcal{W}$ of a vector space $\mathcal{Y}$ over $\mathbb{F}$ such that $\mathcal{U}+\mathcal{V}=\mathcal{Y}$, but $\mathcal{W} \neq(\mathcal{W} \cap \mathcal{U}) \dot{+}(\mathcal{W} \cap \mathcal{V})$. [HINT: Simple examples exist with $\mathcal{Y}=\mathbb{R}^2$.]
If $\mathcal{U}_j, j=1, \ldots, k$, are finite dimensional subspaces of a vector space $\mathcal{Y}$ over $\mathbb{F}$, then the sum
(4.10) $\mathcal{U}_1+\cdots+\mathcal{U}_k=\left\{\mathbf{u}_1+\cdots+\mathbf{u}_k: \mathbf{u}_i \in \mathcal{U}_i\right.$ for $\left.i=1, \ldots, k\right\}$
is said to be direct if
$$
\operatorname{dim} \mathcal{U}_1+\cdots+\operatorname{dim} \mathcal{U}_k=\operatorname{dim}\left\{\mathcal{U}_1+\cdots+\mathcal{U}_k\right\} .
$$
If $\mathcal{U}=\mathcal{U}_1+\cdots+\mathcal{U}_k$ and the sum is direct, then we write
$$
\mathcal{U}=\mathcal{U}_1+\cdots+\mathcal{U}_k .
$$
If $k=2$, then formula (2.16) implies that the sum $\mathcal{U}_1+\mathcal{U}_2$ is direct if and only if $\mathcal{U}_1 \cap \mathcal{U}_2=\{0\}$. Therefore, the characterization (4.11) is consistent with the definition of the direct sum of two subspaces given earlier.