# Algebra of Matrices Systems of Linear Equations

ALFONSO SORRENTINO

[Read also § 1.1-7, 2.2,4, 4.1-4]

1. Algebra of Matrices

**Definition. **Let m, n be two positive integers. A m by n real matrix is an
ordered

set of mn elements, placed on m rows and n columns:

We will denote by the
set of m by n real matrices . Obviously, the set

can be identified with R (set of real
numbers).

NOTATION:

Let . To simplify the notation , sometimes we
will abbreviate A =

. The element
is placed on the i-th row and j-th column,
it will be also

denoted by . Morerover, we will denote by
the i-th row
of

A and by the j-th column
of A. Therefore:

**Remark.** For any integer n ≥1, we can consider
the cartesian product R^{n}. Obviously,

there exists a bijection between R^{n} and or
.

**Definition. **Let .
The transpose of A is a matrix

such that:

for any i = 1, . . . , n and j = 1, . .
. , m.

The matrix B will be denoted by A^{T} . In few words, the i-th row of A is simply
the

i-th column of A^{T} :

and .

Moreover,

We will see that the space of matrices can be endowed with
a structure of vector

space. Let us start, by defining two operations in the set of matrices
:
the

addition and the scalar multiplication.

• Addition:

where A + B is such that:
.

• Scalar multiplication:

where cA is such that: .

**Proposition 1.** (, ·) is a vector space. Moreover its dimension is mn.

**Proof. **First of all, let us observe that the zero vector is the zero matrix 0
[i.e., the

matrix with all entries equal to zero: 0_{ij} = 0]; the opposite element (with
respect

to the addition) of a matrix A, will be a matrix −A, such that
.

We leave to the reader to complete the exercise of verifying that all the axioms
that

define a vector space hold in this setting.

Let us compute its dimension. It suffices to determine a basis. Let us consider
the

following mn matrices (for h = 1, . . . ,m and k = 1, . . . , n), such that:

[in other words, the only non- zero element of
is the
one on the h-th row and

k-th column]. It is easy to verify that for any
, we have:

therefore this set of mn matrices is a spanning set of . Let us verify that

they are also linearly independent . In fact, if

This shows that is a basis.

**Definition.** Let .

• A is a square matrix (of order n), if m = n; the n-ple
is
called

diagonal of A. The set will be simply denoted by
.

• A square matrix is said upper triangular [resp. lower

triangular] if a_{ij} = 0, for all i > j [resp. if a_{ij} = 0, for all i < j].

• A square matrix is called diagonal if it is both upper and lower

triangular [i.e., the only non-zero elements are on the diagonal: a_{ij} = 0,

for all i ≠ j].

• A diagonal matrix is called scalar, if
.

• The scalar matrix with is called unit matrix,

and will be denoted I_{n}.

• A square matrix is symmetric if A = A^{T} [therefore, a_{ij} = a_{ji}]

and it is skew-symmetric if A = −A^{T} [therefore, a_{ij} = −a_{ji}].

Now we come to an important question: how do we multiply two matrices ?

The first step is defining the product between a row and a column vector.

**Definition.** Let and
. We

define the multiplication between A and B by:

More generally, if and , we define the matrix product :

where , for all i = 1, . . . ,m and j = 1, . . . , p.

**Remark. **Observe that this product makes sense only if the number of columns of

A is the same as the number of rows of B. Obviously, the product is always
defined

when the two matrices A and B are square and have the same order.

Let us see some properties of these operations (the proof of which, is left as
an

exercise).

**Proposition 2. **[Exercise]

i) The matrix multiplication is associative; namely:

(AB)C = A(BC)

for any ,
and
.

ii) The following properties hold:

(A + B)C = AC + BC, for any and
;

A(B + C) = AB + AC, for any and
;

, for any
;

(cA)B = c(AB), for any c ∈ R and
,;

(A + B)^{T} = A^{T} + B^{T} , for any ;

(AB)^{T} = B^{T}A^{T} , for any and
.

**
Remark.** Given a square matrix
, it is not true in general that there

exists a matrix such that AB = BA = I

_{n}. In case it does, we say that

A is invertible and we denote its inverse by A

^{-1}.

Let us consider in the subset of invertible matrices:

: there exists such that AB = BA = I

_{n}} .

This set is called general linear group of order n.

We leave as an exercise to the reader, to verify that the following properties hold.

**Proposition 3**. [Exercise]

i) For any , we have: .

ii) For any , we have: and .

iii) For any and c ∈ R, with c ≠ 0, we have: . In

particular, .

An important subset of , that we will use later on, is the set of orthogonal

matrices.

**Definition. **A matrix
is called orthogonal, if:

The set of the orthogonal matrices of order n is denoted
by O_{n}(R). Moreover,

from what observed above (i.e., A^{-1} = A^{T} ), it follows that
(i.e.,

orthogonal matrices are invertible).

To conclude this section, let us work out a simple exercise, that will provide
us with

a characterization of O_{2}(R).

**Example. **The set O_{2}(R) consists of all matrices of the form:

for all α ∈ R.

**
Proof. **One can easily verify (with a direct computation), that:

therefore, these matrices are orthogonal. We need to show
that all orthogonal

matrices are of this form.

Consider a matrix . Let us show that there
exists α ∈ R,

such that or
. By Definition:

From the first two equations , if follows that b^{2} = c^{2}. There
are two cases:

i) b = c or ii) b = −c .

i) In this case, plugging into the other equations, one gets that (a + d)c = 0

and therefore:

i') c = 0 or i'') a + d = 0 .

In the case i'):

namely, A is one of the following matrices:

In the case i ):

therefore,
with α∈ [0, 2π ) such that (a, b) = (cos^{2}α,
sin^{2}α) (this

is possible since a^{2} + b^{2} = 1).

ii) Proceeding as above, we get that (a − d)c = 0 and
therefore there are two

possibilities:

ii') c = 0 or ii'') a − d = 0 .

In the case ii'), we obtain again:

In the case ii ):

therefore, with α∈ [0, 2π ) such that (a, c) = (cos^{2}α,
sin^{2}α) (this

is possible since a^{2} + c^{2} = 1).

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