2 votes

A base of the orthogonal complement of F = Cl {(2,1,2,2)} is?

To solve it I am converting it into an equation of 4 unknowns in the following way; 2x + y + 2z + 2w = 0. After this I clear x to get the vectors that would come to be the base, but it does not give me the expected result which is; {(-1,2,0,0), (- 1,0,1,0), (- 1,0,0,1)}. They could explain to me what I should do.


Anon User Points 0
To solve for x is x=-y/2 -z-w We are left then (-y/2 -z-w,y,z,w). That can be separated into (-1/2,1,0,0)and+(-1,0,1,0)z+(-1,0,0,1)w. It gives us as a basis {(-1/2,1,0,0),(-1,0,1,0),(-1,0,0,1)}. The issue is that those three bases generate this subspace. And you can rewrite the vectors that are other that are linearly dependent. (-1/2,1,0,0)=2(-1,2,0,0). This basis can be written then as {(-1,2,0,0),(-1,0,1,0),(-1,0,0,1)}. The issue is that if you write well or write as the above, both represent the same subspace.


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