Consider the subspaces of $\mathrm{R^{20}}$, $ W_1= \{(x_1,..., x_{20}) \in \mathrm{R^{20}} : x_i = 0 \, \text{when}\, i | 20\}$
$W_2= \{(x_1,..., x_{20}) \in \mathrm{R^{20}} : x_i = 0\, \text{when} \,i= 4k | 20 ,\, k = 1,2,..,5 \}$ Then find $\dim(W_1 + W_2)$
My approach :
$\dim(W_1) = 20- 6 =14$
$\dim(W_2) = 20- 2 =18$.
Now I'm confused about number of elements in $\dim (W_1 \cap W_2)$, should it be $14$, since $x_i = 0$ for $6$ elements in intersection .
We have that $W_1 \subset W_2$, hence
$W_1 \cap W_2= W_1$ and $W_1+W_2=W_2.$