I tried solving the following linear algebra problem, I hope that someone can tell me if this is a good solution, and if not, how I should solve it.
Let $U$ and $W$ be vector subspaces of the vector space $\mathbb R^3$:
$U=\{(a,b,c):a=b=c\}$ and $W=\{(a,b,c):a=0\}$
Prove that $\mathbb R^3=U\oplus W$.
Now, I know the following rule: the sum of subspaces $U$ and $W$ is direct if and only if every vector $x\in U+W$ can be represented uniquely as $x = u+w$ where $u \in U$ and $w \in W$.
I also know that the sum of two vector spaces is direct if their sum is trivial.
I think my solution is incomplete because I only showed that the sum of $U$ and $W$ is direct, so can someone help me with the rest? Here it goes:
Let $\forall(a,b,c) \in \mathbb R$ and $\exists b', b'', c', c''$ such that:
$(a,b,c) = (a, b'', c'')+(0, b', c')$
where $(a, b'', c'') \in U$ and $(0, b', c') \in W$
Let's see if that sum is direct (unique for every vector $(a,b,c)$):
$(5,6,8) = (5, 5, 5) + (0, 1, 3)$
There is no other way to represent this vector since vector $u = (5, 5, 5)$ is already predetermined by the first coordinate of the vector $(5,6,8)$. What is true for this vector, follows for all others.
Now, did I just prove only that $U+W$ is a direct sum, or did I also prove that it's $\mathbb R^3$ as well? Because at the beginning, I took the vector $\forall(a,b,c) \in \mathbb R$.
Best Answer
Well, it seems like you actually have a much easier proof on your hands: W is the (y,z) plane and therefore contains the vectors (0,1,0) and (0,0,1). U contains the vector (1,1,1). These three vectors are linearly independent, so you have $\mathbb{R}^3$ if you choose them as your basis.