Linear algebra proof - Orthogonal complements

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SUMMARY

The discussion centers on proving that for an inner product space V and a finite dimensional subspace W, if a vector x is not in W, there exists a vector y in V that lies in the orthogonal complement of W, such that the inner product of x and y is non-zero. The theorem referenced states that any vector y in V can be uniquely expressed as the sum of a vector u in W and a vector z in the orthogonal complement of W. The conclusion drawn is that since x is not in W, the corresponding component u must be non-zero, leading to a non-zero inner product with y.

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  • Understanding of inner product spaces
  • Knowledge of orthogonal complements
  • Familiarity with finite dimensional vector spaces
  • Basic grasp of vector decomposition in linear algebra
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  • Study the properties of inner product spaces
  • Learn about the concept of orthogonal complements in linear algebra
  • Explore vector decomposition techniques in finite dimensional spaces
  • Review theorems related to projections in inner product spaces
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Homework Statement



Let V be an inner product space, and let W be a finite dimensional subspace of V. If x is not an element of W, prove that there exists y in V such that y is in the orthogonal complement of W, but the inner product of x and y is not equal to 0.

Homework Equations


The Attempt at a Solution



I'm pretty lost. There is a theorem which states:

Let W be a finite dimensional subspace of an inner product space V, and let y be in V. Then there exist unique vectors u in W andf z in the orthogonal complement of W such that y=u+z.

But I don't know how this would apply.
 
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You can use that. Call U the orthogonal complement of W. Then your vector can be split into x=w+u where w is in W and u is in U. If x is not in W then u is not zero, agree with that? The inner product of u with u is then nonzero. Do you see it now?
 

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