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

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

Here's a hint:

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.


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