How do I find an orthonormal basis for a plane defined as $x_1+x_2+x_3=0$? I don't know where to start -- not even given dimensions. Thank you.
Find an orthonormal basis of a plane
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linear-algebra
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0Do you mean a plane in $\mathbb{R}^3$ given as $ax+by+cz=0$? – 2011-03-04
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1In general: find an arbitrary basis, and apply Gram-Schmidt. – 2011-03-04
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0@Arturo: I am not given dimensions, which is part of my confusion. – 2011-03-04
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0What does "a+b+c=0" mean, then? Are a, b, and c constants? Variables? Do you know the dimension of the ambient space? – 2011-03-04
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0@Arturo: a, b, and c are variables (it was given by x_1 x_2 and x_3 in the textbook). Nothing stated about the dimensions. – 2011-03-04
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1@John: You should stick to $x_1$, $x_2$, and $x_3$. Those are *clearly* variables, whereas $a$, $b$, and $c$ are usually constants or scalars. – 2011-03-04
1 Answers
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The subspace in $\mathbb{R}^n$ defined by $x_1+x_2+x_3=0$ is of dimension $n-1$ and has a basis consisting of the vectors \begin{align*} &(1,0,-1,0,0,\ldots,0)\\ &(0,1,-1,0,0,\ldots,0)\\ &(0,0,0,1,0,\ldots,0)\\ &(0,0,0,0,1,\ldots,0)\\ &\qquad\qquad\vdots\\ &(0,0,0,0,0,\ldots,1) \end{align*} The last $n-3$ vectors are already normal, pairwise orthogonal, and orthogonal to the first two. So all you need to do is orthogonalize $(1,0,-1,0,\ldots,0)$ and $(0,1,-1,0,\ldots,0)$ by using Gram-Schmidt, and then normalize.
The dimension of the ambient space does not matter in the end, except in so far as padding the basis.