Exercise 1.6.29

then we have

is contained in both $W_1$ and $W_2$ and hence in $W_1\cap W_2$. But if $v\ne 0$ and can be express as $u={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k{a}_i^{\prime }{u}_i$, then we have ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^m{b}_i{v}_i-{\sum }_{i=1}^k{a}_i^{\prime }{u}_i=0$. This is contradictory to that $\{u_1,\dots ,v_1,\dots \}$ is a basis of $W_1$. Hence we have

this means $a_i=b_j=c_l=0$ for all index $i$, $j$, and $k$. So the set $\beta =\{u_1,\dots ,v_1,\dots ,w_1,\dots \}$ is linearly independent. Furthermore, for every $x+y\in W_1+W_2$ with $x\in W_1$ and $y\in W_2$ we can find the representation $x={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k{d}_i{u}_i+{\sum }_{i=1}^m{b}_i{v}_i$ and $y={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k{d}_i^{\prime }{u}_i+{\sum }_{i=1}^n{c}_i{w}_i$. Hence we have

is linear combination of $\beta$. Finally we have dim$(W_1+W_2)=k+m+n=$dim$(W_1)+$dim$(W_2)-$dim$(W_1\cap W_2)$ and hence $W_1+W_2$ is finite-dimensional.

if and only if dim$(W_1\cap W_2)=0$. And dim$(W_1\cap W_2)=0$ if and only if $W_1\cap W_2=\{0\}$. And this is the sufficient and necessary condition for $V=W_1\oplus W_2$.