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 , some times 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 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 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 rations -in-excel.html">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
fol lowing 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^TA^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 O₂(R).

Example. The set O₂(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 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
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) = (cos²α, sin²α) (this
is possible since a² + c² = 1).