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Algebra of Matrices Systems of Linear Equations


[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).


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 Rn. Obviously,
there exists a bijection between Rn 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 AT . In few words, the i-th row of A is simply the
i-th column of AT :

and .


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: 0ij = 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 aij = 0, for all i > j [resp. if aij = 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: aij = 0,
for all i ≠ j].
• A diagonal matrix is called scalar, if .
• The scalar matrix with is called unit matrix,
and will be denoted In.
• A square matrix is symmetric if A = AT [therefore, aij = aji]
and it is skew-symmetric if A = −AT [therefore, aij = −aji].

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

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 = AT + BT , for any ;
(AB)T = BTAT , for any and .

Given a square matrix , it is not true in general that there
exists a matrix such that AB = BA = In. 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 = In} .

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

Definition. A matrix is called orthogonal, if:

The set of the orthogonal matrices of order n is denoted by On(R). Moreover,
from what observed above (i.e., A-1 = AT ), 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 O2(R).

Example. The set O2(R) consists of all matrices of the form:

for all α ∈ R.

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 b2 = c2. 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) = (cos2α, sin2α) (this
is possible since a2 + b2 = 1).

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

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) = (cos2α, sin2α) (this
is possible since a2 + c2 = 1).


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