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## #1 2007-04-19 09:06:58

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

### Linear algebra problem

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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## #2 2007-04-19 09:13:04

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

### Re: Linear algebra problem

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

which is perpendicular to the hyperplane V.

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

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## #3 2007-04-19 09:39:32

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

### Re: Linear algebra problem

This is how I might go about it.

Since V is not the whole space

, there is a vector a in
which is not in V. This is not the zero vector since 0 must be in the subspace V. ∴ ||av|| > 0 for all vV. Now consider the following set (of positive real numbers):

This 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 ||av[sub]0[/sub]|| = u.
(iii) x·v = 0 for all vV, where x = av[sub]0[/sub].
(iv) For all w

, if x·w = 0, then wV.

If you can prove these claims, then thats it. However, they may not be true. Im 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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## #4 2007-04-19 12:26:40

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

### Re: Linear algebra problem

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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