The hyperplane $H$ defined by $$H:=\{x\in\mathbb {R}^n:a^Tx=b\}$$ is the set that has dimension $n-1$, my question is why or how can we prove that its dimension is $n-1$? Thank you to every one who provide any help or if possible the proof for that.
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$$\begin{eqnarray*}(a_1 a_2 \cdots a_n) \left(\begin{array} {c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array}\right) &=& b\\a_1x_1 + a_2x_2 + \cdots + a_n x_n &=& b\\x_n&=&\frac{1}{a_n}\left(b-a_1x_1-a_2x_2-\cdots -a_{n-1}x_{n-1}\right)\end{eqnarray*} $$ So $x\in H$ if and only if it has the form $$\left(\begin{array} {c} x_1 \\ x_2 \\ \vdots \\x_{n-1}\\ \frac{1}{a_n}\left(b-a_1x_1-a_2x_2-\cdots -a_{n-1}x_{n-1}\right) \end{array}\right)$$ Now, this is in the span of which vectors? What is the basis?
Here
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Alexander Gruber
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It is going to be in the span of $x_{1}, x_{2},..., x_{n-1}$ right? – rinan yosf Dec 16 '14 at 05:26
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Not quite, you need vectors, those xs are scalars – Alexander Gruber Dec 16 '14 at 05:38
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We have: $a^Tx = b\iff a_1x_1+a_2x_2+\cdots + a_nx_n = b$. Let $x_1, x_2,\cdots , x_{n-1}$ be any real numbers. Then $x_n$ is uniquely determined. You can write the solution in term of $x_1, x_2,\cdots, x_{n-1}$. This means the dimension is $n-1$.
DeepSea
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Thanks but I think any affine subspace can be written as $a_1x_1+a_2x_2+\cdots + a_nx_n = b$ and not necessary to be a hyperplane so how is that going to be its dimension? – rinan yosf Dec 16 '14 at 05:31
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I left some details out, but the complete answer would be the same as that given by the first answer above. – DeepSea Dec 16 '14 at 05:32
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I mean how we are going to discriminate the affine subspace of dimension $n-2$ for instance from a hyperplane using the same way of defining the affine subspace above? I am sorry for asking maybe silly questions but I am really trying to understand these concepts and topics – rinan yosf Dec 16 '14 at 05:38
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Hints:
(1) If $\;V_{\Bbb F}\;$ is any vector space and $\;0\neq f\in V^*\;$ is any (non-zero) linear functional, then $\;f\;$ is always onto.
(2) If $\;0\neq f\in V^*\;$ , then $\;\ker f\;$ is always a hyperplane (= a maximal proper subspace of $\;V\;$ )
(3) If $\;\dim_{\Bbb F}V=n<\infty\;$ , then with the same notation and assumptions as above $\;\dim\ker f=n-1\;$
Timbuc
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@Hoot , indeed so, but in fact $;H=\ker f+b;$ , for some $;0\neq f\in V^*;$ . By definition, it dimension is $;\dim\ker f;$ . – Timbuc Dec 16 '14 at 04:38
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First of all, I think the way you frame the question is confusing. A hyperplane is n-1 dimensional by definition. You should probably be asking "How to prove that this set- Definition of the set H goes here- is a hyperplane, specifically, how to prove it's n-1 dimensional"
With that being said. The proof can be separated in two parts:
-First part (easy): Prove that H is a "Linear Variety"
- Second Part: Prove that this linear variety H is n-1 dimensional.
I hope the answer from this question would be sufficient
Help with proof: Hyperplane is an $(n-1)$-dimensional linear variety
Hussin
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