Direct sum of vector subspaces equals $\mathbb R^3$

Question

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 $\forall(a,b,c) \in \mathbb R$.

Answer 1

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.

Answer 2

You need to prove the existence of the decomposition to show the sum cover $\mathbb{R}^3$ and its uniqueness to show the sum is direct.

Existence Let $v = (x,y,z) \in \mathbb{R}^3$. Then, let $u = (x,x,x) \in U$ and let $w = (0,y-x,z-x) \in W$. Indeed, $v = u + w$.

Uniqueness Let $v = u + w$ and $v = u' + w'$. Then, $u + w = u' + w'$ which means that $u - u' = w' - w \in W$. But obviously $u - u' \in U$, thus $u-u' \in U \cap W$. But the only element of the intersection is $0$ (you can check yourself easily). Thus $u = u'$ and it follows that $w = w'$.

Answer 3

There is a nice theorem which states that you have a decomposition of V into a direct sum of subspaces $W_i \subset V$ if and only if there exist projection maps $\pi_1:V \rightarrow V$, $\pi_2:V \rightarrow V$ with $\pi_1 \circ \pi_2 = \pi_2 \circ \pi_1 = 0$, $\pi_1 + \pi_2 = id_V$ and $Im(\pi_i) = W_i$. So in your case, I would define

$$ \pi_1\begin{pmatrix}a\\b\\c\end{pmatrix} := \begin{pmatrix}a\\a\\a\end{pmatrix}$$

$$ \pi_2\begin{pmatrix}a\\b\\c\end{pmatrix} := \begin{pmatrix}0\\b-a\\c-a\end{pmatrix} $$

As you can see, the two mappings composed give the zero map for any $v$, their sum gives the identity and their images are the two subspaces desired.

Answer 4

$U=\{(a,b,c):a=b=c\}$ and $W=\{(a,b,c):a=0\}$

You can use this: $$\dim(U)+ \dim(W) = \dim(U\cap W) + \dim(U+W)$$ Easy calculation shows that $\dim(U) = 1$ and $\dim(W)=2$, and $U\cap W = \{(0,0,0)\}$. Hence, $\dim(U\cap W) = 0$ and $$1+2=0+\dim(U+W) \\ \implies \dim(U+W) = 3$$ Can you take it from here?