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Let $U⊆ℝ^4$ a subspace given by the equations:

$$x_2+2x_3=0 \mbox{ and }x_2+x_4=0$$

Find an orthonormal basis of $U^⊥$.

I think these two vectors are in $U^⊥$: $(0,1,2,0)^T,(0,1,0,1)^T$

How can I quickly justify that these are there is no other third vector in the basis of $U^⊥$.

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It's because $U^{\bot}$ has dimension 2. Is that clear? –  Nils Matthes Dec 18 '12 at 18:21
    
No, how can I see that? Because there are two equations? –  Kasper Dec 18 '12 at 18:23
1  
$U$ has dimension 2, because it is the intersection of two distinct hyperplanes in $\mathbb{R}^4$(use the dimension formula). Hence its orthogonal complement $U^{\bot}$ must have dimension 2 as well. –  Nils Matthes Dec 18 '12 at 18:25
    
Ah, I understand, thanks ! –  Kasper Dec 18 '12 at 18:30
    
Then try writing it up and post it as an answer. This will be the acid test for your understanding of the problem. When you can do this, then you've understood the problem and its solution. –  Nils Matthes Dec 18 '12 at 18:34
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1 Answer 1

up vote 2 down vote accepted

Orthonormal bases require unit vectors that are pairwise orthogonal. Both of your proposed basis vectors are too long, and are not orthogonal to each other. In fact, your second vector doesn't even lie in $U^\perp$. (Why?)

Note that $$\begin{align}U &= \bigl\{(x_1,x_2,x_3,x_4)^T\in\Bbb R^4:x_2+2x_3=0,x_2+x_4=0\bigr\}\\ &= \bigl\{(x_1,x_2,x_3,x_4)^T\in\Bbb R^4:x_2=-2x_3,x_4=-x_2\bigr\}\\ &= \bigl\{(x_1,-2x_3,x_3,2x_3)^T:x_1,x_3\in\Bbb R\bigr\}\\ &= \bigl\{x_1(1,0,0,0)^T+x_3(0,-2,1,2)^T:x_1,x_3\in\Bbb R\bigr\}.\end{align}$$

Since $U$ is spanned by the linearly independent vectors $(1,0,0,0)^T$ and $(0,-2,1,2)^T$, then $U$ has dimension $2$. If we let $S:\Bbb R^4\to\Bbb R^4$ be the orthogonal projection onto $U$, then $U$ is the image (range) of $S$, and $U^\perp$ is the kernel (null space) of $S$. By Dimension Theorem, we have $$4=\dim(\Bbb R^4)=\dim(U)+\dim(U^\perp)=2+\dim(U^\perp),$$ so $\dim(U^\perp)=2$. That's how we know that there are $2$ vectors in any (orthonormal) basis for $U^\perp$.

To actually find an orthonormal basis for $U^\perp$, we start by finding a basis for $U^\perp$, then performing the Gram-Schmidt process. Note (by our characterization of $U$ above) that $(y_1,y_2,y_3,y_4)^T\in U^\perp$ if and only if the following chain of equivalent conditions holds for all real $x_1,x_3$ (hereafter $\cdot$ is the standard inner product, or dot product): $$0=(x_1,-2x_3,x_3,2x_3)^T\cdot(y_1,y_2,y_3,y_4)^T$$ $$0=x_1y_1-2x_3y_2+x_3y_3+2x_3y_4$$ $$0=x_1y_1+x_3(-2y_2+y_3+2y_4)$$

Since the above equation must hold for $x_3=0$ and $x_1\neq 0$, then we must have $y_1=0$ in order for $(y_1,y_2,y_3,y_4)^T\in U^\perp$. (Why?) We'll also need $y_3=2y_2-2y_4$ for $(y_1,y_2,y_3,y_4)^T\in U^\perp$. (Why?) These two conditions are both necessary and sufficient, so $$\begin{align}U^\perp &= \bigl\{(y_1,y_2,y_3,y_4)^T\in\Bbb R^4:y_1=0,y_3=2y_2-2y_4\bigr\}\\ &= \bigl\{(0,y_2,2y_2-2y_4,y_4)^T:y_2,y_4\in\Bbb R\bigr\}\\ &= \bigl\{y_2(0,1,2,0)^T+y_4(0,0,-2,1)^T:y_2,y_4\in\Bbb R\bigr\}.\end{align}$$

This gives us the basis $$\bigl\{(0,1,2,0)^T,\,(0,0,-2,1)^T\bigr\}$$ for $U^T$. Can you perform the Gram-Schmidt process on this basis and take it from here?

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