Productive matrix

Short description: Type of matrix in linear algebra

In linear algebra, a square nonnegative matrix A of order n is said to be productive, or to be a Leontief matrix, if there exists a n×1 nonnegative column matrix P such as P−AP is a positive matrix.

## Contents

* 1 History
* 2 Explicit definition
* 3 Properties
* 3.1 Characterization
* 3.2 Transposition
* 4 Application
* 5 References

## History

The concept of productive matrix was developed by the economist Wassily Leontief (Nobel Prize in Economics in 1973) in order to model and analyze the relations between the different sectors of an economy.[1] The interdependency linkages between the latter can be examined by the input-output model with empirical data.

## Explicit definition

The matrix A∈Mn,n(ℝ) is productive if and only if A⩾0 and ∃P∈Mn,1(ℝ),P>0 such as P−AP>0.

Here Mr,c(ℝ) denotes the set of r×c matrices of real numbers, whereas >0 and ⩾0 indicates a positive and a nonnegative matrix, respectively.

## Properties

The following properties are proven e.g. in the textbook (Michel 1984).[2]

### Characterization

Theorem A nonnegative matrix A∈Mn,n(ℝ) is productive if and only if In−A is invertible with a nonnegative inverse, where In denotes the n×n identity matrix.

Proof

"If" :

Let In−A be invertible with a nonnegative inverse,
Let U∈Mn,1(ℝ) be an arbitrary column matrix with U>0.
Then the matrix P=(In−A)−1U is nonnegative since it is the product of two nonnegative matrices.
Moreover, P−AP=(In−A)P=(In−A)(In−A)−1U=U>0.
Therefore A is productive.

"Only if" :

Let A be productive, let P>0 such that V=P−AP>0.
The proof proceeds by reductio ad absurdum.
First, assume for contradiction In−A is singular.

The endomorphism canonically associated with In−A can not be injective by singularity of the matrix.
Thus some non-zero column matrix Z∈Mn,1(ℝ) exists such that (In−A)Z=0.
The matrix −Z has the same properties as Z, therefore we can choose Z as an element of the kernel with at least one positive entry.
Hence c=supi∈[|1,n|]zipi is nonnegative and reached with at least one value k∈[|1,n|].
By definition of V and of Z, we can infer that:

cvk=c(pk−∑i=1nakipi)=cpk−∑i=1nakicpi
cpk=zk=∑i=1nakizi, using that Z=AZ by construction.
Thus cvk=∑i=1naki(zi−cpi)≤ 0, using that zi≤cpi by definition of c.
This contradicts c>0 and vk>0, hence In−A is necessarily invertible.
Second, assume for contradiction In−A is invertible but with at least one negative entry in its inverse.

Hence ∃X∈Mn,1(ℝ),X⩾0 such that there is at least one negative entry in Y=(In−A)−1X.
Then c=supi∈[|1,n|]−yipi is positive and reached with at least one value k∈[|1,n|].
By definition of V and of X, we can infer that:

cvk=c(pk−∑i=1nakipi)=−yk−∑i=1nakicpi
xk=yk−∑i=1nakiyi, using that X=(In−A)Y by construction
cvk+xk=−∑i=1naki(cpi+yi)⩾0 using that −yi⩽cpi by definition of c.
Thus xk≤−cvk<0, contradicting X⩾0.
Therefore (In−A)−1 is necessarily nonnegative.

### Transposition

Proposition The transpose of a productive matrix is productive.

Proof

Let A∈Mn,n(ℝ) a productive matrix.
Then (In−A)−1 exists and is nonnegative.
Yet (In−AT)−1=((In−A)T)−1=((In−A)−1)T
Hence (In−AT) is invertible with a nonnegative inverse.
Therefore AT is productive.

## Application

With a matrix approach of the input-output model, the consumption matrix is productive if it is economically viable and if the latter and the demand vector are nonnegative.

## References

1. ↑ Kim Minju, Leontief Input-Output Model (Application of Linear Algebra to Economics)
2. ↑ Philippe Michel, "9.2 Matrices productives", Cours de Mathématiques pour Economistes, Édition Economica, 1984

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