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**quackensack****Member**- Registered: 2007-02-27
- Posts: 47

This problem was confusing me so any explanations would be helpful and greatly appreciated!

Let V be a subspace of R^n with dim(v) = n - 1. (Such a subspace is called a hyperplane in R^n.) Prove that there is a nonzero x

R^n such that V = {v R^n|x v = 0}.*Last edited by quackensack (2007-04-19 09:11:30)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

In other words, you need to show that there exists a nonzero vector **x** in

*Last edited by JaneFairfax (2007-04-19 09:14:04)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

This is how I might go about it.

Since V is not the whole space

, there is a vectorThis set is bounded below and so (by the completeness of the real numbers) it has a greatest lower bound *u*. Now I make the following claims:

(i) *u* > 0.

(ii) There exists some **v**[sub]0[/sub] ∈ *V* such that ||**a**−**v**[sub]0[/sub]|| = *u*.

(iii) **x**·**v** = 0 for all **v** ∈ *V*, where **x** = **a**−**v**[sub]0[/sub].

(iv) For all **w** ∈

If you can prove these claims, then thats it. However, they may not be true. Im only making these claims based on my visualization of the problem in

, so I feel that they are intuitively reasonable claims; nevertheless I could be mistaken in my intuition.*Last edited by JaneFairfax (2007-04-19 09:42:22)*

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**whatismath****Member**- Registered: 2007-04-10
- Posts: 19

I think the assertion to be proved is:

there is a nonzero x in R^n, such that x.v = 0 for all v in V.

Actually, the statement is true for any finite-dimenisional

inner product space U (say dim U = n).

Since V is not the entire U and 0 is in V, there is nonzero x in U\V.

Let {v_1, v_2, ..., v_(n-1)} be a normalized orthogonal base of V.

Consider w = x - (x.v_1)v_1 - (x.v_2)v_2 - ... - (x.v(n-1))v_(n-1).

Then w is nonzero, else x is in V.

For any v_i, i = 1, 2, ..., n-1,

w.v_i = x.v_i - (x.v_i)(v_i.v_i) = 0

since v_i.v_i = 1, and v_j.v_i = 0 for i different from j.

For any v in V, v = a_1(v_1) + ... + a_(n-1)(v_(n-1))

for some scalar (R here) a_1, ..., a_n-1.

So w.v = a_1(w.v_1) + ... + a_(n-1)(w.v_(n-1)) = 0.

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