Let $U,V$ two subspaces of $\mathbb{R}^n$, such that:
$U = \{x \in \mathbb R^n | x_1+x_2+…+x_n=0\}$ and $V=\{x\in V |x_1=x_2=…=x_n\}$
Show that the sum of the two subspaces is a direct sum.
What I did:
We want to show that $U\cap V = \{0\}$.
Indeed, let $w \in U\cap V$, Hence:
$w_1 = w_2 = … = w_n$
$w_1 + w_2 + … + w_n = 0.$
it's easy to see that $\forall i.w_i = 0$, Therefore it's the zero vector.
Now, for my understanding this isn't suffice in order to show it's a direct sum.
We also need to prove that $U+V = \mathbb{R}^n$.
Is it right? How to do that?
Best Answer
You don't have to do that. The task is to prove that $U+V$ is a direct sum, that is what you showed. As far as you wrote, you don't have to prove that $U+V = \mathbb R$.
If you do want to prove that, look at the dimension of $U$ and $V$. Because $U\cap V = 0$, you know that dim$(U+V) = $ dim$(U)+$dim$(V)$.
If you do not yet have the theoretical background to make the dimension argument, then take ant $x\in\mathbb R^n$ and look at $\tilde x$ defined as $\tilde x_i =x_i - \bar x$ where $\bar x$ is the average value of values $x_1,x_2,\dots, x_n$.