In linear algebra, a square nonnegative matrix A {\displaystyle A} of order n {\displaystyle n} is said to be productive, or to be a Leontief matrix, if there exists a n × 1 {\displaystyle n\times 1} nonnegative column matrix P {\displaystyle P} such as P − A P {\displaystyle P-AP} is a positive matrix.
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. The interdependency linkages between the latter can be examined by the input-output model with empirical data.
Explicit definition The matrix A ∈ M n , n ( R ) {\displaystyle A\in \mathrm {M} _{n,n}(\mathbb {R} )} is productive if and only if A ⩾ 0 {\displaystyle A\geqslant 0} and ∃ P ∈ M n , 1 ( R ) , P > 0 {\displaystyle \exists P\in \mathrm {M} _{n,1}(\mathbb {R} ),P>0} such as P − A P > 0 {\displaystyle P-AP>0} . Here M r , c ( R ) {\displaystyle \mathrm {M} _{r,c}(\mathbb {R} )} denotes the set of r×c matrices of real numbers, whereas > 0 {\displaystyle >0} and ⩾ 0 {\displaystyle \geqslant 0} indicates a positive and a nonnegative matrix, respectively.
Properties The following properties are proven e.g. in the textbook (Michel 1984).
Characterization Theorem A nonnegative matrix A ∈ M n , n ( R ) {\displaystyle A\in \mathrm {M} _{n,n}(\mathbb {R} )} is productive if and only if I n − A {\displaystyle I_{n}-A} is invertible with a nonnegative inverse, where I n {\displaystyle I_{n}} denotes the n × n {\displaystyle n\times n} identity matrix. Proof "If" :
Let I n − A {\displaystyle I_{n}-A} be invertible with a nonnegative inverse, Let U ∈ M n , 1 ( R ) {\displaystyle U\in \mathrm {M} _{n,1}(\mathbb {R} )} be an arbitrary column matrix with U > 0 {\displaystyle U>0} . Then the matrix P = ( I n − A ) − 1 U {\displaystyle P=(I_{n}-A)^{-1}U} is nonnegative since it is the product of two nonnegative matrices. Moreover, P − A P = ( I n − A ) P = ( I n − A ) ( I n − A ) − 1 U = U > 0 {\displaystyle P-AP=(I_{n}-A)P=(I_{n}-A)(I_{n}-A)^{-1}U=U>0} . Therefore A {\displaystyle A} is productive. "Only if" :
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