# Matrix Operations

**Theorem**. If the columns of C each sum to less than 1, and the entries

in
are non-negative, then (I
- C) is invertible, and the production vector

has nonnegative entries and is the unique solution to

Expanding the problem.

If the external demand is
, the various sectors can produce
units, but this

will create an intermediate demand of
.

They can then produce another
units. This leads to further intermediate

demand of
units.

They can then produce another
units. This leads to further intermediate

demand of
units.

Adding up all of the above, in total the industries have to produce this to
satisfy

the external final demand as well as all of the internal demands that arise as a

result.

Now for a trick to see this sum in another way. Let

W = I + C + C^{2} + C^{3} + . . . + C^{m}

Then

and so

W - CW = (I - C)W =

If the columns of C each sum to values strictly less than 1, then
(an

n × n matrix of zeros) as
. And by the

Theorem, (I - C) is invertible, and so

which implies

I + C + C^{2} + . . . + C^{m} ≈ (I - C)^{-1}

if the column sums of C are less than 1. We can make the approximation as

accurate as desired by making m sufficiently large.

In practice, this can be a useful approximation, especially if the matrix is
large

and sparse (i.e., contains many zeros) and the column sums are significantly

less than one. Even a standard linear system
can be rewritten to use

this trick, by letting C = I -A. Then we can write the system as

**Economic meaning of entries in (I - C) ^{-1}.**

Given external demand
, say we have found the solution (production vector)

satisfying

Now consider a new external demand
Say we have solution

satisfying

If we let
then

because everything is linear.

Now say
Then we have solution
But this is just

the _______________________________of (I -C)^{-1} ! So if we increase external

demand from
up to
, we need to increase production from

to
,
where
.

**Result:**The j^{th} column of (I - C)^{-1} says how much extra production will

be needed from all sectors, to meet one additional unit of demand for sector j.

Because the production vector
contains the production levels for each
sector,

and because
, we can go even further, and get.

**Result :**The number in row i, column j of (I - C)^{-1} says how much extra

production will be required of industry i in order to meet an increase in exter-

nal demand of one unit of sector j, taking into account the new intermediate

demands which arise.

This could also be written as the partial derivative

## Section 2.8 { Subspaces, Basis for Null Space and Column Space

Note. examples labeled with Arabic numbers are examples from the book, examples labeled with Roman

numerals are not.

**Subspaces**

Definition Given a set W of vectors (such as R^{n}), a subspace of W is any

subset
satisfying three properties.

a. The zero vector is in H

b. If
and
are in H, then
. (We say H is closed under addition).

c. If, then for all scalar constants c, the vector
. (We say H is

closed under scalar multiplication). Note. this includes c = 0 and c < 0.

**Example 1. **span of two vectors.

Say
, and H = Spanf. Then H is a subspace of R^{n}.

Check.

a.
is in the span of
and
, so
.

b. If
and
, it means
and
for some

constants r_{1}, r_{2}, s_{1}, and s_{2}. Then,

,
which is in H since it is a linear combination of

and
(and therefore in the span of
and
).

c. If
, it means
for some constants r_{1} and r_{2}. Then,

which is in H.

**Example 2. **Line L in R^{2} not through the origin.

**Example III**. The set

. (Read the colon as
"such
that" or "satisfying".) Check the three properties.

**Example IV. **The set

Definition. The column space of a matrix A (written "Col A") is the set of

all linear combinations of the columns of A, i.e. the span of the columns of A.

Definition. The null space of a matrix A (written "Nul A") is the set of all

solutions to the homogeneous equation

**Theorem 12:** For an m × n matrix A, the set H =Nul A is a subspace of R^{n}.

**Proof: **Note. a vector
is in
by definition of this H.

Definition. A basis for a subspace H is a linearly independent set in H that

spans H.

**Example V:** The vectors
are a basis for R^{n}. In fact,
they

are called the standard basis.

**Example 6:** Find a basis for the null space of the matrix

Write the solution to
in parametric vector form.

So x_{2}, x_{4}, and x_{5} are free,

So Nul A =_____________________________

And is a linearly independent set. So in fact it is a basis for Nul A.

This always happens, because.

1. When we write the solution set to
in parametric vector form, the

solution set is always the set of all linear combinations of the vectors
attached

to the free variables. That is, those vectors always span Nul A.

2. For a general matrix A and the equation
, if x_{i} is a free variable,

think about the i^{th} position in each of the vectors in the parametric vector
form

of the solution set to
. That position will be 1 in the vector which

goes with x_{i}, and will be 0 in all other vectors. The same holds for the other

free variables. So the only linear combination of those vectors which gives

will be the trivial linear combination, i.e. the vectors will always be linearly

independent.

Result 1:To find a basis for the null space of A, write the solution set toin parametric vector form. The vectors in that form are a basis. |

Finding a basis for the column space of A takes less work, but some explanation

for why the technique works.

**Example 7: **Find a basis for the column space of the matrix

Calling the columns
note that

So any linear combination

can really be written as

So in fact
spans
Col B. Also, the set
is linearly indepen-

dent because those vectors are in fact the vectors
, and
which
is

clearly a linearly independent set.

Since
, this also means that
. This in
turn

implies that

Now consider the matrix

If you check, you will find that A is row equivalent to B on the previous page.

In fact, B = rref(A). The fact that A and B are row equivalent means that

This in turn means that
where
are the
columns

of A. So the same tricks we used with B will work for A, and in fact

spans Col A.

Also
is a linearly independent set (you still need to give it a
little

bit of thought to fill in the details to see this).

Result 2: (Theorem 13) To find a basis for the column space of A, reduce Ato an echelon form to find out where the pivots are (you don't need reduced echelon form). The pivot columns of A form a basis for the column space of A. (Warning: be sure you use the columns of the original A once you find out where the pivots are, and not the columns of the echelon form of A!!!) |