Orthonormal basis for subsets of C^3

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SUMMARY

The discussion focuses on finding an orthonormal basis for the subspace W, defined as W=span({(i,0,1)}) in C^3. It is established that a vector x in W_perpendicular satisfies the condition = 0, leading to the conclusion that c = -ai. The participants clarify that the norm of w1 is zero, indicating that W does not have a valid basis and is essentially the origin, which implies that W_perpendicular encompasses all of C^3. The importance of using the complex conjugate in the inner product for complex spaces is also emphasized.

PREREQUISITES
  • Understanding of complex vector spaces, specifically C^3.
  • Knowledge of inner product definitions, particularly for complex vectors.
  • Familiarity with the concept of orthonormal bases and their properties.
  • Basic linear algebra concepts, including spans and perpendicular subspaces.
NEXT STEPS
  • Study the properties of complex inner products in vector spaces.
  • Learn about the implications of zero-length vectors in linear algebra.
  • Explore the concept of orthonormal bases in higher-dimensional complex spaces.
  • Investigate the relationship between spans and their orthogonal complements in C^n.
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Mathematicians, physicists, and students studying linear algebra or complex vector spaces, particularly those interested in orthonormal bases and their applications in higher dimensions.

gysush
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We want to find a basis for W and W_perpendicular for W=span({(i,0,1)}) =Span({w1}) in C^3

a vector x =(a,b,c) in W_perp satisfies <w1,x> = 0 => ai + c = 0 => c=-ai
Thus a vector x in W_perp is x = (a,b,-ai)

So an orthonormal basis in W would be simply w1/norm(w1) ...but the norm(w1)=0 (i^2 + 1 = 0)

What am I missing here? Does a basis for W satisfy that it has zero length? Thus it is just the origin. Then would all of C^3 be W_perp?
 
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The complex inner product norm <x,y> is defined by (x*)^T y. You are forgetting the complex conjugate.
 
Ahh...I forgot to remember that a norm for F=C requires we take the complex conjugate of the 2nd vector.

You beat me to to it. :-)
 

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