I have two subspaces $V_1$, $V_2$ of $\mathbb{R}^n$ each specified as the span of a given list L1, L2 of basis vectors, and I know a priori that $V_1\subset V_2$. What is the quickest (or shortest) expression to calculate a basis of the orthogonal complement of $V_1$ inside $V_2$?
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2 Answers
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Generally $V_{1}^{\perp} \cap V_{2} = \left( V_{1}+ V_{2}^{\perp}\right)^{\perp}$ holds, so this can be easily achieved with the following code:
NullSpace[Union[L1,NullSpace[L2]]]
Example:
n=6;
dim1=2;
dim2=5;
L2=RandomInteger[{-9,9},{dim2,n}]
(*{{-8,-8,-9,7,8,-5},
{-4,6,7,-7,3,-4},
{7,9,-2,5,7,-3},
{-8,6,-2,-1,4,4},
{-8,6,-1,-5,-7,7}}*)
L1=RandomInteger[{-3,3},{dim1,dim2}].L2
(*{{-43,-7,-11,-8,-2,3},{-45,-37,-12,-10,-23,9}}*)
basis=NullSpace[Union[L1,NullSpace[L2]]]
(*{{3423303,26965449,6851892,0,0,137110328},
{-8348533,-97751035,69911228,0,137110328,0},
{4717317,-1179017,-42619288,34277582,0,0}}*)
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The function "Orthogonalize" calculates a Gram-Schmidt base of its arguments. Assume that the argument is: vs={v1,v2,..}, then "Orthogonalize" first normalizes vs[[1]]: e1. Then it calculates a unit vector e2 from e1 and vs[[2]] perpendicular to e1. Then a unit vector e3 from e1, e2 and vs[[3]] perpendicular to e1 and e2, e.t.c
Here is an example: Given a subspace V2 of dimension 4 of R^5 by:
bas2=RandomReal[{-1, 1}, {4, 5}];
Then a subbase V1 ofV2 of dimension 2 can be specified by 2 linear combinations of bas2:
bas1 = RandomReal[{-1, 1}, {2, 4}] . b2;
If we apply "Orthogonalize" to "Join¨bas1,bas2" we get 6 vectors, where the first 2 are a orthonormal base for V1 and the 3. and 4. are an orthonormal base of the complement and the last 2 are zero because "Join¨bas1,bas2" is linear depended.
combas = Orthogonalize[Join[b1, b2]];
ortb1= combas[[;;2]]
ortcompl= combas[[3;;4]]
L1andL2, together with the expected output. $\endgroup$