# Leontief Model of Production
## Setting

Suppose there is an economy with $n$ goods. For each $i = 1, \dots , n,$ there is a single producer of good $i$. To output one unit of good $i$, the producer requires $a_{1i}$ units of good $1$, $a_{2i}$ units of good $2$, etc.

We assume that $a_{ji} \ge 0$ for all $i$ and $j$. Let $x_i$ be the amount of good $i$ produced in the economy and $c_i$ the amount of good $i$ consumed (not used as inputs). The market for good $i$ clears (supply equals demand) when the following linear equation holds.

$$x_i = a_{i1} x_1 + a_{i2} x_2 + \dots + a_{in} x_n + c_i$$

The production matrix is the matrix $A$ of required inputs for each good $i$ and the consumption vector $c$ is the vector of consumption of each good $i$.

$A = \begin{pmatrix} a_{11} & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ . & . &... &.\\ a_{n1} & a_{n2} & \dots & a_{nn}\end{pmatrix}$, $x = \begin{pmatrix} x_1 \\ x_2 \\... \\ x_n\end{pmatrix}$, and $c = \begin{pmatrix} c_1 \\ c_2 \\... \\ c_n\end{pmatrix}$

Let $I_n$ be the identiy matrix of order $n$, we have the following equation,

$$(I_n - A)x = c.$$

## Productive

The economy is called productive if there is a non-negative solution to the system of equations that captures market clearing.

## Theorem
If the sum of the entries in row $i$ of $A$ is less than $1$ for all $i = 1, \dots , n$ goods, then the economy is productive.

## Preliminary Proof

### Invertible

First, we want to show that $I_n -A$ is invertible.

Suppose not, there exists a non-zero vector $x \in R^n$ such that $(I_n - A )x = x - Ax =0$. Let $x_i$ be the element of $x$ with the largest absolute value, and $A_i$ be the corresponding row vector of $A$. In other words, $x_i - A_i x=0$, where $0$ is a scalar. However, the sum of $A_i$ is smaller than $1$, so it is impossible. $I_n -A$ is invertible.

### Non-negative solution

Second, we need to show that all solution is non-negative. That is, for any non-zero vector $y \in R^n$ and $(I_n - A )x = y$, $x$ must be a non-zero vector in $R^n_+$.

Suppose not, there is a non-zero $y \in R^n_+$ and $(I_n - A )x = y$, and at leat one element of $x$ is negative. Let $x_i$ be the negative element of $x$ with the largest absolute value, $A_i$ be the corresponding row vector of $A$, and $y_i$ be the corresponding element of $y$. We have $x_i - A_i x=y_i$. Again, the elements of $A_i$ are non-negative and the sum is less than $1$. Hence, $x_i - A_i x$ could not be positive, which is a contractdition.