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Consider $\mathbb R^4$ together with the standard inner product and let $W=\{(x_1, x_2, x_3, x_4) \in \mathbb R^4 |\ x_1+x_2-x_4=0\}$. Find a basis for $W^\perp$.

How do you solve this question? All I can think of is that you have to put the elements in $W$ in terms of others.

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  • $\begingroup$ No need to download any piece of software. Just write your question in LaTeX. For example, \$x_1\$ to get $x_1$, \$W^\perp\$ to get $W^\perp$. $\endgroup$
    – Siminore
    Jun 15, 2012 at 14:23

5 Answers 5

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This is not the most elegant way, but here are some hints.

(1) First note that $W$ is the vector space given by $$W = \{(t, s, u, t+s)\in \mathbb{R}^4 \lvert t,s,u \in \mathbb{R} \}. $$

(2) Then for a vector $(a,b,c,d)$ to be in $W^{\perp}$ we need $$(a,b,c,d)\cdot (t,s,u,t+s) = 0 $$ for all $t,s,u \in \mathbb{R}$.

(3) We get from this that

$$W^{\perp} = \{(a, a, 0, -a) \lvert a \in \mathbb{R}\}. $$

(4) From here you should be able to write down a basis.

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Since this is a hw, I will only give some hint!

First find a basis for $W$. To do so use the identity $x_4=x_1+x_2$ that defines $W$. You should find that any member $x=(x_1,x_2,x_3,x_4)$ of $W$ is a linear combination of 3 linearly independent vectors $w_1,w_2,w_3$, i.e. $w=\alpha_1 w_1+\alpha_2 w_2+\alpha_3w_3$ for some real numbers $\alpha_1,\alpha_2, \alpha_3$ (e.g. depending on $x_1,x_2,x_3$). Next, find a non-zero vector $v=(v_1,v_2,v_3,v_4)$ which is orthogonal to $w_1, \ w_2$, and $w_3$, meaning that $$ v\cdot w_1=v\cdot w_2= v\cdot w_3=0. $$ Any vector of the form $\alpha.v$, where $\alpha \in \mathbf{R}\setminus\{0\}$ is a basis for $W^\perp$.

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Consider that $(1,1,0,-1)$ is orthogonal to all $(x_1,x_2,x_3,x_4)\in W$ because by definition $x_1+x_2-x_4=0$. Then $W^{\perp}=\langle(1,1,0,-1)\rangle$.

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  • $\begingroup$ I still don't know what to do.. So how would I find the basis using the $W^\perp$? $\endgroup$
    – Alice
    Jun 15, 2012 at 14:41
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Note that $W = \{(1,1,0,-1)\}^\perp =\langle(1,1,0,-1)\rangle^\perp$. So $W^\perp = \big(\langle(1,1,0,-1)\rangle^\perp\big)^\perp= \langle(1,1,0,-1)\rangle$, with basis $\{(1,1,0,-1)\}$.

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I would first find a basis for the solution space to the homogenous system of equations given by $x_1 + x_2 - x_4 = 0$. This turns out to be a 3 dimensional vector space. To find a basis for the solution space, set $x_2 = 1, x_3 = 0$, and $x_4 = 0$, then solve for $x_1$. This will provide you with your first vector in the basis. For the second vector, repeat this process, again solving for $x_1$, but setting $x_3 = 1$ instead of $x_2$. For the third and final vector, set $x_4 = 1$ and solve for $x_1$. You can then set up a new system of equations where the vector $\langle y_1, y_2, y_3, y_4 \rangle$ inner-producted with each of the basis vectors you just found will be zero. This will give you 3 equations in 4 variables. The solution space to this new system will provide you with a basis for $W^{\perp}$.

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  • $\begingroup$ I only found one basis vector for the homogenous system $x_1 +x_2 -x_4 =0$ which is (0, 0, 0, 1). Am I doing this wrong? All I did was put it in row reduced form but this is what I got.. $\endgroup$
    – Alice
    Jun 15, 2012 at 15:14
  • $\begingroup$ What happens if you set $x_4 = 0, x_3 = 1,$ $and x_2 = 0$ and then solve for $x_1$? So far, you've found one of the vectors in the basis for the solution set;) Wait, my bad... (0,0,0,1) does not satisfy the equation. You have to change one of the numbers (but not the $x_3$). $\endgroup$
    – Andrew
    Jun 15, 2012 at 15:15
  • $\begingroup$ @Alice: Just updated the answer with some more detail. $\endgroup$
    – Andrew
    Jun 15, 2012 at 15:36

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