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October 20, 2014

October 20, 2014

Posted by **Simon** on Sunday, November 20, 2011 at 3:09pm.

- Linear Algebra -
**MathMate**, Sunday, November 20, 2011 at 11:47pmThe dimension of vector space R3 is 3.

n=(2,3,2) occupies one of the three dimensions, so the subspace orthogonal to n has a dimension of two, i.e. two vectors span the remaining subspace.

The basis of the remaining subspace can be in many forms. One way is to start with an arbitrary vector, say a1=(1,1,0) and apply Gram-Schmidt process to transform it into a vector orthogonal to n. After normalization, we get A1=(7,2,-10)/(3sqrt(17)).

Note that A1.n=0.

Similarly, if we start with a2=(1,0,1), and apply Gram-Schmidt process, we get A2=(1,0,-1)/sqrt(2).

Note also that A2.n=0, and A1.A2=0.

Thus we have a basis for R3 {n,A1,A2} of which two are orthogonal to n. (n being one of the three vectors of the basis).

- Linear Algebra -
**MathMate**, Monday, November 21, 2011 at 12:28am... and the basis for the subspace orthogonal to n is {A1,A2}.

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