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There are three industries interrelated so that their outputs are used as inputs by themselves.

$$ A \;\; =\;\; [a_{jk}] \;\; = \;\; \left [ \begin{array}{ccc} 0.1 & 0.5 & 0 \\ 0.8 & 0 & 0.4 \\ 0.1 & 0.5 & 0.6 \\ \end{array} \right ] $$

Here

  1. $a_{jk}$ is the fraction of the output of industry $k$ consumed by industry $j$.
  2. $p_j$ be the price charged by industry $j$ for its total output.

A problem is to find prices so that for each industry, total expenditures equal total income. Show that this leads to $Ap = p$, where $p = \left [ \begin{array}{ccc} p_1& p_2 & p_3\\ \end{array} \right ]$, and find a solution $p$ with nonnegative $p_1, p_2, p_3$. Also, show that the consumption matrix $A$ must have columns which sum to $1$ and always has eigenvalue $1$. What approach should I follow to solve this question.?

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Solving the equation $Ap = p$, you are looking for an eigenvector corresponding to the eigenvalue $\lambda = 1$, so the null space of $A - I$ should provide you with adequate vectors. To prove that $1$ is always an eigenvalue, first realize that $\det A = \det A^T$, which leads to the fact that $A$ and $A^T$ have the same eigenvalues. It is easily proven for a left stochastic matrix (note that these follow the exact same form as your consumption matrix) such as $A$ that the following vector $\vec{u} =\begin{bmatrix}1 \\1 \\1 \end{bmatrix}$ is always an eigenvector of $A^T$ corresponding to $\lambda = 1$ (see below). Consequently, $1$ must also be an eigenvalue of $A$. $$A^T\begin{bmatrix}1 \\1 \\1 \end{bmatrix}=\begin{bmatrix}0.1 & 0.8 &0.1 \\0.5 & 0 & 0.5 \\0.6 & 0.4 &0 \end{bmatrix}\begin{bmatrix}1 \\1 \\1 \end{bmatrix} = \begin{bmatrix}1 \\1 \\1 \end{bmatrix}$$

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  • $\begingroup$ Thanks @Hyperion. $\endgroup$
    – nbalodi
    Jun 4, 2019 at 2:15

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