<p>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.</p>

<p>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</p>

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

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

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

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

<p>Set $(1) = (2)$.</p>

<p>$\implies 1 - p_1 = 1 \implies p_1 = 0$</p>

<p>which is a contradiction.</p>

CC BY-SA 3.0
2017-08-17T21:16:09.880Z
2017-08-17T21:16:09.880Z
Kitsune Cavalry
<p>Let matrix $A = \begin{bmatrix} p_1 &amp; p_2 &amp; \ldots &amp; p_n \end{bmatrix}$. Let $\mathbf{x}^*$ be a fixed solution to $A \mathbf{x} = c$. Then for any vector $\mathbf{u}$ that belongs to the <a href="https://en.wikipedia.org/wiki/Kernel_(linear_algebra)" rel="noreferrer">null space</a> 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 $A$.</p>

<p>By the <a href="https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem" rel="noreferrer">rank-nullity theorem</a>, the <a href="https://en.wikipedia.org/wiki/Rank_(linear_algebra)" rel="noreferrer">rank</a> of matrix $A$ plus the nullity (dimensionality of the <a href="https://en.wikipedia.org/wiki/Kernel_(linear_algebra)" rel="noreferrer">null space</a>) of matrix $A$ equals the number of columns of $A$:</p>

<p>$$ \operatorname{rank}(A) + \operatorname{nullity}(A) = n$$</p>

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

CC BY-SA 3.0
2017-08-17T20:55:49.747Z
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)
Kitsune Cavalry
CC BY-SA 3.0
2017-08-17T15:43:28.307Z
2017-08-17T15:43:28.307Z
Matthew Gunn
<blockquote>
  <p>Take the set of all vectors $x = (x_1, \cdots, x_n)$ that are
  solutions to $p_1x_1 + \cdots + p_nx_n = I &gt; 0$. Show that this set
  has $n-1$ dimensions.</p>
</blockquote>

<p>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.</p>

<p>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 $\{x_1 \cdots x_{n-1}\}$:</p>

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

<p>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.</p>

<p>So I note that $p_1x_1 + \cdots + p_{n-1}x_{n-1} = I - p_nx_n$, and that $I &gt; 0$.</p>

<p>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.</p>

CC BY-SA 3.0
2017-08-17T15:45:02.073Z
I glanced at your question quickly, but perhaps [proofs of the rank-nullity theorem](https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem#Proofs) have the argument you're looking for?
Matthew Gunn
<mathematical-economics><linear-algebra>
Budget hyperplane in n dimensions
Kitsune Cavalry
# Budget hyperplane in n dimensions

(Asked by: username_0 on 2017-08-17)

> 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.

**Tags:** mathematical-economics, linear-algebra

> **Comments:**
> **username_1**: I glanced at your question quickly, but perhaps [proofs of the rank-nullity theorem](https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem#Proofs) have the argument you're looking for?  

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## 2 answers

### Answer by username_1 on 2017-08-17. Score: 5 | Accepted answer

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](https://en.wikipedia.org/wiki/Kernel_(linear_algebra)) 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](https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem), the [rank](https://en.wikipedia.org/wiki/Rank_(linear_algebra)) of matrix $A$ plus the nullity (dimensionality of the [null space](https://en.wikipedia.org/wiki/Kernel_(linear_algebra))) 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 $n-1$.

> **Comments:**
> **username_0**: 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)  

---

### Answer by username_0 on 2017-08-17. Score: 1

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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