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Very confused on how to deal with these direct sum problems.

Problem:Suppose $U=\{(x,y,x+y,x-y,2x) \in \mathbb{F}^{5}:x,y \in \mathbb{F}\}$

Find a subspace $W$ of $\mathbb{F}^{5}$ such that $\mathbb{F}^{5}=U \oplus W$

Trying to figure out a routine way to do these problems. I used the following link to help Find a subspace $W$ of $\mathbb{F}^4$ such that $\mathbb{F}^4 = U \oplus W$

Attempt:

Given $(a,b,c,d,e) \in \mathbb{F}^{5}$,

$(a,b,c,d,e)=(a,b,a+b+c-a-b,a-b-a+b+d,2a-2a+e)$

$=(a,b,a+b,a-b,2a)+(0,0,c-a-b,-a+b+d,e-2a)$

where $(a,b,a+b,a-b,2a) \in U$ and $(0,0,c-a-b,-a+b+d,e-2a) \in W$

Hence $\mathbb{F}^{5}=U+W$

Next Show $U \cap W=\{0\}$

Attempt:

Let $(e,f,g,h,i) \in U \cap W$ then $e=0,f=0$

I can't seem to figure out why $g=0,h=0,i=0$

Also is this the correct way to approach this type of problem?

Thanks

2 Answers2

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Since $U=\text{span}\left(\left\{\right(1,0,1,1,2), \, (0,1,1,-1,0)\}\right)$ so to find a $W$, one can choose $W=U^{\perp}$ (orthogonal complement of $U$), i.e. $$W=\{v \in \Bbb{F}^5 \, | \, \forall u \in U, \,\, v \cdot u = 0\}.$$ In this particular problem, (using the basis vectors of $U$) $$W=\{(x,y,z,s,t) \, | \, x+z+s+2t=0 \text{ and } y+z-s=0\}.$$ Thus we need the basis vectors for the solution set of \begin{align*} x+ z+s+2t & =0\\ y+z-s & =0. \end{align*} The solutions are given by $$W=\{(-z-s-2t,-z+s,z,s,t) \, | \, z,s,t \in \Bbb{F}\}.$$ OR $$\begin{bmatrix}x\\y\\z\\s\\t\end{bmatrix}=z\begin{bmatrix}-1\\-1\\1\\0\\0\end{bmatrix}+s\begin{bmatrix}-1\\1\\0\\1\\0\end{bmatrix}+t\begin{bmatrix}-2\\0\\0\\0\\1\end{bmatrix}.$$ $$W=\text{Span}\left(\left\{(-1,-1,1,0,0), \, (-1,1,0,1,1),\, \, (-2,0,0,0,1)\right\}\right).$$

Since $W=U^{\perp}$, so $U \cap W=\{0\}$ is an easy outcome of that.

Anurag A
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I don't see your answer panning out.

Here's another method:

$U$ is the span of $\{(1,0,1,1,2), (0,1,1,-1,0)\}$. (To see this, plug in $x=1,y=0$, and $x=0, y=1$, to get two obviously l.i. vectors in $U$. But $U$ is clearly two-dimensional: two free variables.)

So the problem can be boiled down to expanding this to a basis of $\Bbb F^5$.

You could use the "sifting algorithm", applied to a generating set (most easily just adjoin the standard basis to the two vectors above to get a basis). See this answer.

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    Would I also have to say $W={(0,0,w,z,s) \in \mathbb{F}^{5}:w,z,s \in \mathbb{F}}$ using different variables from $x$ and $y$?Then is the problem finished? –  Dec 20 '19 at 03:46
  • Let me look at it some more. I'll get back to you. –  Dec 20 '19 at 03:50
  • My first attempt was wrong. I don't see where yours is going. I have provided an alternate route, if you're still interested. –  Dec 20 '19 at 06:46
  • I would like to see the alternative route –  Dec 20 '19 at 17:21
  • It's straight forward. Did you look at the link? –  Dec 20 '19 at 17:31
  • Now I see it. My bad –  Dec 20 '19 at 17:33
  • If I'm not mistaken, you can adjoin ${e_1,e_2, e_5}$ to get a basis. –  Dec 20 '19 at 17:41