1

The question is the same as one previously asked, but I can't comment so I had to ask my own. Find a subspace $W$ of $\mathbb{F}^4$ such that $\mathbb{F}^4 = U \oplus W$

Suppose $U=\{ (u_1,u_1,u_2,u_2)\in \mathbb{F}^4:u_1,u_2\in \mathbb{F}\}$.
Find a subspace $W$ of $\mathbb{F}^4$ such that $\mathbb{F}^4 = U\oplus W$.

I chose $W = \{ (0,w_1,0,w_2)\in \mathbb{F}^4:w_1,w_2\in \mathbb{F}\}$.
Is this incorrect or can I use it just as previous answers have been $(0,w_1,w_2,0)$?

From what I can tell, my answer allows $U\cap W = \{0\}$ and $U+W=\{(u_1,u_1+w_1,w_1,u_2+w_2)\in\mathbb{F}^4:u_1,u_2,w_1,w_2\in\mathbb{F}\}=\mathbb{F}^4$.

Is this a sufficient argument?

Community
  • 1
joewfu16
  • 13

2 Answers2

0

It is true that $U\oplus W=\Bbb{F}^4$. Indeed $U\cap W=\{0\}$, but the space $U+W$ consists of all sums of vectors from $U$ and $W$. That is \begin{eqnarray*} U+W&=&\{u+w:\ u\in U,\ w\in W\}\\ &=&\{(u_1,u_1,u_2,u_2)+(0,w_1,0,w_2):\ u_1,u_2\in\Bbb{F},\ w_1,w_2\in\Bbb{F}\}\\ &=&\{(u_1,u_1+w_1,u_2,u_2+w_2):\ u_1,u_2,w_1,w_2\in\Bbb{F}\}. \end{eqnarray*} So, loosely speaking, you shouldn't only look at sums of vectors with the same $x$ and $y$, but at all pairs of vectors from $U$ and $W$.

Servaes
  • 63,261
  • 7
  • 75
  • 163
  • 1
    Thanks. I am going to grad school this fall, taking linear algebra 2 without having had an undergrad course in the field. I am using text and doing exercises to be prepared. I think I see that you mean that my $x,y\in\mathbb{F}$ are not arbitrary enough given the two subspaces and you have named them differently. Should it be $(u_1,u_1+w_1,u_2,u_2+w_2)$ in the last line of the equation array? – joewfu16 Jun 13 '16 at 16:00
  • Oh yes, absolutely! – Servaes Jun 13 '16 at 16:06
  • Trying now to find a subspace of $\mathbb{F}^5$ such that $\mathbb{F}^5=U\oplus W$ when $U={(u_1,u_2,u_1+u_2,u_1-u_2,2u_1)\in \mathbb{F}^5:u_1,u_2\in\mathbb{F}}$.

    I have $W={(0,0,w_1,w_1,w_1)\in\mathbb{F}^5:w_1\in\mathbb{F}}$.

     – joewfu16 Jun 13 '16 at 18:10
0

The example you provide doesn't work. The vector $(1,1,1,1)\in \mathbb{F}^4$ but can't be written as $(x,2x,y,2y)$. But you can still solve your problem by taking the orthogonal complement $W^{\bot}$ of $W$. With $W$ as defined, its orthogonal complement is given by:$$W^{\bot}=\left\lbrace X=(x,y,z,t)\in\mathbb{F}^4\mid\,\langle X,v\rangle=0,\mbox{ for all }v\in W \right\rbrace ,$$ where $\langle \cdot,\cdot\rangle$ represents the usual inner product of $\mathbb{F}^4$.

On the orther hand, we can write $$W=\mbox{span} \left\lbrace \begin{pmatrix} 1\\ 1\\ 0\\ 0 \end{pmatrix} , \begin{pmatrix} 0\\ 0\\ 1\\ 1 \end{pmatrix} \right\rbrace .$$ Therefore $X=(x,y,z,t)\in W^{\bot}$ if and only if: \begin{align*} \left\{ \begin{array}{l} \overline{x}=-\overline{y}\\ \overline{z}=-\overline{t} \end{array} \right. \end{align*} This system provides $$W^{\bot}=\mbox{span} \left\lbrace \begin{pmatrix} -1\\ 1\\ 0\\ 0 \end{pmatrix} , \begin{pmatrix} 0\\ 0\\ -1\\ 1 \end{pmatrix} \right\rbrace .$$ You can immediately check that $W\oplus W^{\bot}=\mathbb{F}^4$.

Tighana Basele
  • 1
  • I am only in the first chapter of the book and know nothing of spans or orthogonal complements, so what you have shown is a little foreign to me still. Does my edit of the example I have make a difference? – joewfu16 Jun 13 '16 at 18:54