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# 實分析一題

If M={x:L(x)=0},Where L is a continuous linear functional on H,prove

that M^(倒T) is a vector space of dimension 1(unless H不等於M)

H is a Hilbert space

### 2 Answers

- densityLv 61 decade agoFavorite Answer
Proof : Let u be a nonzero vector perpendicular to M. Multiplying it by a constant, you can assume that L(u) = 1.

Let v be any other vector perpendicular to M. We have to show that v is a constant multiple of u.

In fact, v= L(v)u. Here is how to see it.

Let w = v-L(v)u.

Then w is in M because L(w) = 0. (use linearity of L and the fact that L(u)=1).

On the other hand, w is perpendicular to M, because the perpendicular space is a subspace and u and v are both in it.

So w = 0, because the only vector in both M and M perp is the zero vector.

- ?Lv 41 decade ago
令 N = {y ∈ H: 〈x, y〉= 0 for all x ∈ M}.

怪怪的英文，需要改善一下，沒有人會叫 N 做 perpendicular space，

有正式的名稱，叫做 M 的 orthogonal complement。

在 Hilbert space，我還真沒有看過有人用 perpendicular 來形容垂直，都是用 orthogonal 的，這是相對於一個 inner product 來用。