Calculus subspace

Question

Let L be spanned by 2,-5,-1,2 Find a basis of the orthogonal complement L⊥ of L.

Answer 1

You're given that the subspace L⊂R4 is spanned by the vector:

v=(2,−5,−1,2) You are asked to find a basis for the orthogonal complement L⊥, which consists of all vectors in R4 that are orthogonal to v.

Here are the steps:

We want to find all vectors x=(x1,x2,x3,x4)∈R4 such that:

x⋅v=0

That is:

2x1−5x2−x3+2x4=0

This is a single linear equation in four variables, so the solution set will be a subspace of dimension 3, and we can find a basis with 3 vectors.

We solve:

2x1−5x2−x3+2x4=0

Let's solve for x1:

x1=(5/2)x2+(1/2)x3−x4

​Now write the general solution as a linear combination:

Let:

1. x2=a
2. x3=b
3. x4=c

Then:

x1=(5/2)a+(1/2)a=b−c

So the general vector x∈L⊥ is:

x=((5/2)a+(1/2)a=b−c, a, b, c)

We can now extract a basis by writing this as a linear combination of independent vectors:

x=a(5/2,1,0,0)+b(1/2,0,1,0)+c(−1,0,0,1)

To simplify, multiply the fractional vectors by 2 to eliminate denominators:

=a⋅(5,2,0,0)+b⋅(1,0,2,0)+c⋅(−2,0,0,2)

Final Answer: Basis for L⊥

A basis for the orthogonal complement L⊥⊂R4 is:

{(5,2,0,0), (1,0,2,0), (−2,0,0,2)}