# Show that $V = W \oplus U$.

The Problem

Let $V=k^3$ for some field $k$. Let $W$ be the subspace spanned by $(1,0,0)$ and let $U$ be the subspace spanned by $(1,1,0)$ and $(0,1,1)$. Show that $V= W \oplus U$. Explain your argument in detail.

What I Know

1. I know that a field $k^n=n$-tuples of elements of $k$.

2. I know that a subset $W$ of a vector space $V$ over a field $k$ is a subspace if the operations of $V$ make $W$ into a vector space over $k$.

3. I know that if span$(S)=V$ for a set $S$ in a vector space $V$, where $S$ is linearly independent, then $S$ is a basis for $V$.

4. I know that the external direct sum $V \oplus W$ for vector spaces $V$ and $W$ over a field $k$ is defined as the set of all ordered pairs $(v,w)$ such that $v\in V$ and $w \in W$.

What I Don't Know

1. How to apply what I listed above to help me solve the problem. I am absolutely atrocious at this material and struggle so much in simply starting these problems.

2. If everything I listed above is even relevant to the problem at hand.

3. If what I listed above is insufficient to complete the problem.

Text: Abstract Linear Algebra by Curtis

• It sounds like you need to know the internal direct sum. – Slade Feb 28 '17 at 11:45
• Alright, but why? – Thy Art is Math Mar 1 '17 at 8:09

Let $V$ be a vector space and $U, W$ subspaces. Define
$$\phi:U\oplus W\to V$$ $$\phi(u, w)=u+w$$
Obviously $\phi$ is linear. When is it an isomorphism? How can you apply it to your case?
Take the span of $W$ as $a(1,0,0) = (a,0,0)$ for $a$ in your field, and the span of $U$ as $b(1,1,0)+c(0,1,1)=(b,b+c,c)$ with $b,c$ in the field. Now take an element from $V$ say it is $(x,y,z)$ and decompose it in sum of an element in $W$ and an element in $U$ as follows: $(x,y,z) = (x+z-x,0,0) + (y-z,y,z)$ and you can see the first element is in $W$ and the second in $U$ so now, if you proof that it is the unique way of decomposing an element of $V$ as sum of an element in each subspace you would prove $V$ is direct sum of $W$ and $V$
An alternative is to proof the intersection of $W$ and $V$ is ${(0,0,0)}$ and that dim ($W$) =1 and dim ($U$)=2 so you have dim ($V$)=dim ($W$) + dim ($U$) and their intersection is ${(0,0,0)}$ this implies the result.