Exercise 2.C.16

Proof. Let $V=U_1\oplus \cdots \oplus U_m$. Because every vector in $V$ can be written as sum $v_1+\cdots +v_m$ where each $v_j\in U_j$ and since each $v_j$ can be written as sum of $u_1+\cdots +u_{\mathrm{dim<msub><mrow>U</mrow><mrow>j</mrow></msub>}<!--\; FUNCTION\; APPLICATION\; -->}$ for some basis $u_1,\dots ,u_{\mathrm{dim<msub><mrow>U</mrow><mrow>j</mrow></msub>}<!--\; FUNCTION\; APPLICATION\; -->}$ of $U_j$, it follows that the list composed of all such bases spans $V$. Hence $V$ is finite-dimensional.

Moreover, if a linear combination of this list equals $0$, then a linear combination of $v_1,\dots ,v_m$ also equals $0$. But, $U_1+\cdots +U_m$ being a direct sum forces each $u_j$ to equal $0$ and, thus, the coefficients of the basis vectors of $U_j$ must also equal $0$, proving that the list is linear independent.

Therefore, this list is a basis of $V$ and its length is $\dim <!--\; FUNCTION\; APPLICATION\; -->U_1+\cdots +\dim <!--\; FUNCTION\; APPLICATION\; -->U_m$, as desired. □