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
−
A
P
is a positive matrix.
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.
The matrix
A
∈
M
n
(
R
)
is productive if and only if
A
⩾
0
and
∃
P
∈
M
n
,
1
(
R
)
,
P
>
0
such as
P
−
A
P
>
0
.
Theorem A nonnegative matrix
A
∈
M
n
(
R
)
is productive if and only if
I
n
−
A
is invertible with a nonnegative inverse.
Demonstration
Direct involvement :
Let
U
∈
M
n
,
1
(
R
)
,
P
>
0
.
Thus the matrix
P
=
(
I
n
−
A
)
−
1
U
is nonnegative as being the product of two nonnegative matrices.
Moreover,
P
−
A
P
=
(
I
n
−
A
)
P
=
(
I
n
−
A
)
(
I
n
−
A
)
−
1
U
=
U
.
Hence
P
−
A
P
>
0
.
Therefore
A
is productive.
Reciprocal involvement :
We shall proceed by
reductio ad absurdum.
Let us assume
∃
P
>
0
such as
V
=
P
−
A
P
>
0
&
I
n
−
A
is singular.
The endomorphism canonically associated with
I
n
−
A
can not be injective by singularity of the matrix.
Thus
∃
Z
∈
M
n
,
1
(
R
)
non zero such as
(
I
n
−
A
)
Z
=
0
.
The matrix
−
Z
verifies the same properties as
Z
, therefore we can choose
Z
as an element of the kernel with at least one positive entry;
Hence
c
=
sup
i
∈
[
|
1
,
n
|
]
z
i
p
i
is nonnegative and reached with at least one value
k
∈
[
|
1
,
n
|
]
.
By definition of
V
and of
Z
, we can infer that:
c
v
k
=
c
(
p
k
−
∑
i
=
1
n
a
k
i
p
i
)
=
c
p
k
−
∑
i
=
1
n
a
k
i
c
p
i
c
p
k
=
z
k
=
∑
i
=
1
n
a
k
i
z
i
Thus
c
v
k
=
∑
i
=
1
n
a
k
i
(
z
j
−
c
p
j
)
≤
0
.
Yet we know that
c
>
0
and that
v
k
>
0
.
Therefore there is a contradiction,
ipso facto
I
n
−
A
is necessarily invertible.
Now let us assume
I
n
−
A
is invertible but with at least one negative entry in its inverse.
Hence
∃
X
∈
M
n
,
1
(
R
)
,
X
⩾
0
such as there is at least one negative entry in
Y
=
(
I
n
−
A
)
−
1
X
.
Then
c
=
sup
i
∈
[
|
1
,
n
|
]
−
y
i
p
i
is positive and reached with at least one value
k
∈
[
|
1
,
n
|
]
.
By definition of
V
and of
X
, we can infer that:
c
v
k
=
c
(
p
k
−
∑
i
=
1
n
a
k
i
p
i
)
=
−
y
k
−
∑
i
=
1
n
a
k
i
c
p
i
x
k
=
y
k
−
∑
i
=
1
n
a
k
i
y
i
c
v
k
+
x
k
=
−
∑
i
=
1
n
a
k
i
(
c
p
i
+
y
i
)
Thus
x
k
≤
−
c
v
k
<
0
since
∀
i
∈
[
|
1
,
n
|
]
,
a
k
i
⩾
0
,
c
p
i
+
y
i
⩾
0
.
Yet we know that
X
⩾
0
.
Therefore there is a contradiction,
ipso facto
(
I
n
−
A
)
−
1
is necessarily nonnegative.
Proposition The transpose of a productive matrix is productive.
Demonstration
Let
A
∈
M
n
(
R
)
a productive matrix.
Then
(
I
n
−
A
)
−
1
exists and is nonnegative.
Yet
(
(
I
n
−
t
A
)
)
−
1
=
(
t
(
I
n
−
A
)
)
−
1
=
t
(
(
I
n
−
A
)
−
1
)
Hence
(
I
n
−
t
A
)
is invertible with a nonnegative inverse.
Therefore
t
A
is productive.
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.
