In general, what is the strategy for proving that a set is the direct sum of two others?
Let's say I have vector space $E = \mathbb R^2$.
We also have subspaces:
F = {$(x_1,x_2,x_3)\in \mathbb R^3; x_3 = 0$} and
G = {$(a,0,a)\in \mathbb R^3; a \in \mathbb R^3$}
I want to prove that E is the direct sum of F and G; here's what I've done:
We can describe $F = ${$(x_1,x_2,0)\in \mathbb R^3; x_1,x_2 \in \mathbb R$}
Thus, $ F + G = ${$(x_1+a, x_2, a)\in \mathbb R^3; x_1,x_2,a \in \mathbb R$}
Or, $F + G = ${$x(1,0,0) + x_2(0,1,0) + a(1,0,1) \in \mathbb R^3; x_1,x_2,a \in \mathbb R^3$}.
(Let's call the family $b = ((1,0,0),(0,1,0),(1,0,1))$)
Thus, we have $F + G = span(b)$. Since every element of $b$ is linearly independent (which I proved on paper but won't bother to copy onto here), and the $dim(b) = dim(\mathbb R^3)$, we know that $b$ is a base. Therefore, $span(b) = \mathbb R^3$.
I feel like, until this point, the work stands, but correct if I'm wrong on anything.
Now, I have to prove that the intersection of F and G is {(0,0,0)}, and I've proved that the direct sums of F and G is E, right? Was that not proved when I found base $b$? Thank you.
Best Answer
For $E=F \oplus G$ you need that $E=F+G$ and that $F \cap G = \{(0,0,0)\}$. To show that the intersection is the zero vector, suppose that $z=(z_1,z_2,z_3) \in F \cap G$. Since $z \in F$, we have $z_3=0$, and since $z \in G$, we have $z_2=0$ and $z_1=z_3=0$. So $z$ is the zero vector, and since $z$ was arbitrary, the intersection of $F$ and $G$ is the zero vector.
I presume you mean $E=\mathbb{R}^3$, by the way?
Edit: To answer your question of "does $b$ being a basis imply that the sum is direct?" No, I don't think it does because as mentioned in the comments, look at for example $\text{span}((1,0))+\text{span}((1,0))=\text{span}((1,0))$. We have that $\{(1,0)\}$ is a basis of $\text{span}((1,0))$ (like $b$ is a basis for $\text{span}(b)=\mathbb{R}^3$ in your case) but the sum is not direct since the intersection is the vector $(1,0)$. Please correct me if I'm wrong though.