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Take the set of all vectors $x = (x_1, \cdots, x_n)$ that are solutions to $p_1x_1 + \cdots + p_nx_n = I > 0$. Show that this set has $n-1$ dimensions.

I have somehow managed to get myself stuck on the last part of this proof it seems. I am not using the fact that this set is a hyperplane and that hyperplanes are $n-1$ dimensions of the space they are in.

It is easy to show that $\{x_1, \cdots x_n\}$ spans the set we are considering, since $\sum p \cdot x$ is a linear combination and all that. However, $x_n$ can be expressed as a linear combination of $\{x_1 \cdots x_{n-1}\}$:

$$x_n = \frac{I - (p_1x_1 + \cdots p_{n-1}x_{n-1})}{p_n}$$

So we can remove $x_n$ from the span and the resulting set still spans. Now we want to show $\{x_1, \cdots n_{n-1}\}$ are linearly independent. That is, if $p_1x_1 + \cdots p_{n-1}x_{n-1} = 0$, all $p_i = 0$. If the set is spanning and linearly independent, then it is a basis. Since it would have $n-1$ vectors, it would be of dimension $n-1$ and we'd be done.

So I note that $p_1x_1 + \cdots + p_{n-1}x_{n-1} = I - p_nx_n$, and that $I > 0$.

So I assume there is a case where $I - p_nx_n = 0$ and where $I - p_nx_n \neq 0$. I am not sure how to finish off this proof, which makes me sad, because I think I'm just missing something obvious. Any assistance would be appreciated.

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Let matrix $A = \begin{bmatrix} p_1 & p_2 & \ldots & p_n \end{bmatrix}$. Let $\mathbf{x}^*$ be a fixed solution to $A \mathbf{x} = c$. Then for any vector $\mathbf{u}$ that belongs to the null space of $A$, we have $A \mathbf{u} = 0$ hence $\mathbf{x} = \mathbf{x}^* + \mathbf{u}$ is also a solution (furthermore, all solutions $\mathbf{x}$ can be written this way). The dimension of the set of solutions to $A\mathbf{x} = c$ is thus the dimension of the null space of $A$.

By the rank-nullity theorem, the rank of matrix $A$ plus the nullity (dimensionality of the null space) of matrix $A$ equals the number of columns of $A$:

$$ \operatorname{rank}(A) + \operatorname{nullity}(A) = n$$

Matrix A is rank 1 (assuming not all $p_i$ are zero) and hence the nullity is $n-1$.

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  • $\begingroup$ I suppose yeah, this is the easy way to do it. I have never used this method before and was looking to just use linear combinations to show my desired result. I now have the answer and will post it (and in like 3 hours I will be able to upvote your answer) $\endgroup$ – Kitsune Cavalry Aug 17 '17 at 20:55
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I got some outside help for the ending of the proof I was attempting. I'll leave this question if by chance someone else finds it useful.

So if we want to show $p_1x_1 + \cdots + p_{n−1}x_{n−1} = 0 \implies p_i = 0 \quad \forall i$, then assume without loss of generality that $p_1 \neq 0$. We have

$$x_1 = \left(-\frac{p_2}{p_1}\right) x_2 + \cdots + \left(-\frac{p_{n-1}}{p_1}\right)x_{n-1}$$

$$\implies x_1 + \left(\frac{p_2}{p_1}\right) x_2 + + \cdots + \left(\frac{p_{n-1}}{p_1}\right)x_{n-1} = 0 \tag1$$

Then subtract $p \cdot x = 0$ from both sides.

$$(x_1 - p_1x_1) + \left(\frac{p_2}{p_1} x_2 - p_2 x_2\right) + \cdots + \left(\frac{p_{n-1}}{p_1} x_{n-1} - p_{n-1}x_{n-1} \right) = 0$$ $$\implies (1 - p_1)x_1 + \left(\frac{p_2}{p_1} - p_2 \right)x_2 + \cdots + \left(\frac{p_{n-1}}{p_1} - p_{n-1} \right)x_{n-1} = 0 \tag2$$

Set $(1) = (2)$.

$\implies 1 - p_1 = 1 \implies p_1 = 0$

which is a contradiction.

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