Productive matrix explained

of order

is said to be productive, or to be a Leontief matrix, if there exists a

n x 1

such as

P-AP

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\inM_n,n(\R)

is productive if and only if

A\geqslant0

and

\existsP\inM_n,1(\R),P>0

such as

P-AP>0

Here

M_r,c(\R)

denotes the set of r×c matrices of real numbers, whereas

and

\geqslant0

indicates a positive and a nonnegative matrix, respectively.

Properties

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

Characterization

TheoremA nonnegative matrix

A\inM_n,n(\R)

is productive if and only if

I_n-A

is invertible with a nonnegative inverse, where

I_n

denotes the

n x n

identity matrix.

Proof

"If" :

Let

I_n-A

be invertible with a nonnegative inverse,

Let

U\inM_n,1(\R)

be an arbitrary column matrix with

U>0

Then the matrix

P=(I_n-A)^-1U

is nonnegative since it is the product of two nonnegative matrices.

Moreover,

P-AP=(I_n-A)P=(I_n-A)(I_n-A)^-1U=U>0

Therefore

is productive."Only if" :

Let

be productive, let

P>0

such that

V=P-AP>0

The proof proceeds by reductio ad absurdum.

First, assume for contradiction

I_n-A

is singular.

The endomorphism canonically associated with

I_n-A

can not be injective by singularity of the matrix.

Thus some non-zero column matrix

Z\inM_n,1(\R)

exists such that

(I_n-A)Z=0

The matrix

-Z

has the same properties as

, therefore we can choose

as an element of the kernel with at least one positive entry.

Hence

c=\sup_i

	z_i
	p_i

is nonnegative and reached with at least one value

k\in[|1,n|]

By definition of

and of

, we can infer that:

cv_k=c(p_k-

	n
\sum
	i=1

a_kip_i)=cp_k-

	n
\sum
	i=1

a_kicp_i

cp_k=z_k=

	n
\sum
	i=1

a_kiz_i

, using that

Z=AZ

by construction.

Thus

cv_k=

	n
\sum
	i=1

a_ki(z_i-cp_i)\leq 0

, using that

z_i\leqcp_i

by definition of

This contradicts

c>0

and

v_k>0

, hence

I_n-A

is necessarily invertible.

Second, assume for contradiction

I_n-A

is invertible but with at least one negative entry in its inverse.

Hence

\existsX\inM_n,1(\R),X\geqslant0

such that there is at least one negative entry in

Y=(I_n-A)^-1X

Then

c=\sup_i-

	y_i
	p_i

is positive and reached with at least one value

k\in[|1,n|]

By definition of

and of

, we can infer that:

cv_k=c(p_k-

	n
\sum
	i=1

a_kip_i)=-y_k

	n
-\sum
	i=1

a_kicp_i

x_k=y_k-

	n
\sum
	i=1

a_kiy_i

, using that

X=(I_n-A)Y

by construction

cv_k+x_k=-

	n
\sum
	i=1

a_ki(cp_i+y_i)\geqslant0

using that

-y_i\leqslantcp_i

by definition of

Thus

x_k\leq-cv_k<0

, contradicting

X\geqslant0

Therefore

(I_n-A)^-1

is necessarily nonnegative.

Transposition

PropositionThe transpose of a productive matrix is productive.

Proof

Let

A\inM_n,n(\R)

a productive matrix.

Then

(I_n-A)^-1

exists and is nonnegative.

Yet

(I_n-A^T)^-1=((I_n-A)^T)^-1=((I_n-A)^-1)^T

Hence

(I_n-A^T)

is invertible with a nonnegative inverse.

Therefore

A^T

is productive.

Application

See main article: article and Input-output analysis.

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.

Notes and References

Kim Minju, Leontief Input-Output Model (Application of Linear Algebra to Economics)
[Philippe Michel (economist)|Philippe Michel]