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Algebra of Matrices Systems of Linear Equations
ALFONSO SORRENTINO
[Read also § 1.17, 2.2,4, 4.14]
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 ith row and jth column,
it will be also
denoted by . Morerover, we will denote by
the ith row
of
A and by the jth 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 ith row of A is simply
the
ith 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 hth row and
kth 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 nple
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 nonzero 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 skewsymmetric 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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