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Suppose $U = \{(x,x,y,y) \in \mathbb{F}^4 : x, y \in \mathbb{F}\}$

Find a subspace W of $\mathbb{F}^4$ such that $\mathbb{F}^4 = U\oplus W$

Attempt: Now from what I understand I would think that an element of $\mathbb{F}^4$ would look like $$(w,x,y,z) \quad\mbox{such that}\quad w,x,y,z \in \mathbb{F}$$

with that being the case I would use a subspace of the form: $$W = (w-x, 0, 0, z-y) \in \mathbb{F}^4 \quad\mbox{such that}\quad w,x,y,z \in \mathbb{F}$$

But as a solution it was given that $$ W = (0,x,y,0). $$

Explanation?

I think I am not fully grasping how the direct sum sets are formed, but I got the idea that it was using an element from each subspace.

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    $\begingroup$ You might like this web site: detexify.kirelabs.org/classify.html $\endgroup$ – ThePortakal May 4 '16 at 22:48
  • $\begingroup$ @ThePortakal "You da real MVP" $\endgroup$ – dc3rd May 4 '16 at 22:54
  • $\begingroup$ Your answer is also correct, though it's strange to use four parameters to write a two-dimensional space: It's much nicer to write $W = \{(a, 0, 0, b)\}$, or $\langle (1, 0, 0, 0), (0, 0, 0, 1) \rangle$. In general, a proper subspace $U$ of a vector space $V$ admits many complementary subspaces $W \geq V$, that is, subspaces such that $U \oplus W = V$. $\endgroup$ – Travis May 4 '16 at 23:07
  • $\begingroup$ See this is what I am not getting. How is it two dimensional? As well in the subspace given as a solution: $W = (0,x,y,0)$ Wouldn't $U \oplus W = (x,x,y,y) + (0,x,y,0) = (x,2x,2y,y)?$ Now how is that a representation of $\mathbb{F}^4$? $\endgroup$ – dc3rd May 4 '16 at 23:31
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    $\begingroup$ Am I in essence just trying to create another subspace that will allow me to get a vector that characterizes all of $\mathbb{F}^4$? $\endgroup$ – dc3rd May 4 '16 at 23:56
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Let $W=\{(0,w,z,0)\in\mathbb{F}^4:w,z\in\mathbb{F}\}$, then we check that such $W$ is the desired subspace, as the following two steps.

  1. Given $(a,b,c,d)\in\mathbb{F}^4$, it is easy to decompose the vector as below. \begin{align} (a,b,c,d) &=(a,b-a+a,c-d+d,d)\\ &=(a,a,d,d)+(0,b-a,c-d,0), \end{align} where $(a,a,d,d)\in U$ and $(0,b-a,c-d,0)\in W$. Hence $\mathbb{F}^4=U+W$.
  2. If $(e,f,g,h)\in U\cap W$, then we have $e=f$, $g=h$, $e=0$, and $h=0$. So $$(e,f,g,h)=(0,0,0,0)$$ and hence $U\cap W=\{(0,0,0,0)\}$.
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$ U $ is a subspace spanned by the linearly independent set $ S = \{ (1, 1, 0, 0), (0, 0, 1, 1) \} $. Therefore, it suffices to pick a subspace $ W $ which is spanned by two vectors such that their adjoinment to $ S $ would not disturb its linear independence. In other words, we need to extend $ S $ to a basis of $ \mathbb{F}^4 $.

It is easy to see that $ (1, 0, 0, 0), (0, 0, 0, 1) \notin U $. Now, we check if the set $ S' $ formed by adjoining these vectors to $ S $ is linearly independent. The standard method is to row reduce the matrix whose columns are elements of $ S $, but a more direct approach works here. Let $ s_i $ denote the elements of $ S' $:

$$ c_1 s_1 + c_2 s_2 + c_3 s_3 + c_4 s_4 = ( c_1 + c_3, c_1, c_2, c_2 + c_4) $$

For the left hand side to equal zero, it is then clear that we must have $ c_i = 0 $ for all coefficients, establishing linear independence of $ S' $. Therefore, $ S' $ is a basis of $ \mathbb{F}^4 $ (by the dimension theorem), and we may take $ W = \textrm{span} \{(1, 0, 0, 0), (0, 0, 0, 1) \} = \{ (x, 0, 0, y) : x, y \in \mathbb{F} \} $.

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