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Part IIShort introduction to linear
algebra
,Introduction
Linear algebra is concerned with the solutions of so-called “systems
of linear equations”. A system of linear equations is a set of linear
equations that are valid simultaneously. The term “linear” refers to
the fact that the equations contain only linear terms of the variables.
It is the simplest form of algebra that you can think of. So, if x is a
variable whose value we would like to calculate from an equation,
then a linear equation in x is one in which only x or multiples ax of x
occur, where a is a so-called scalar. A scalar is a simple number (In-
teger, Real number, etc) that “scales” the variable x. Terms like x2 or
√
x do not occur in a linear equations. Only first-order powers x1 = x
(or y1 = y or z1 = z, etc.) occur in linear equations. To summarize:
in linear equations, only scalars and linear terms of variables (x, y,
z, etc.) occur. However, multiple variables, all occurring in a linear
fashion are allowed. Therefore, also an equation like ax + by = c
where x and y are variables whose values we would like to know,
and a, b, and c, are scalars, is a linear equation. The most general way
of writing a single linear equation in n variables is
n
∑ ai xi = b
i =1
where x1 , x2 , . . . , xn are the variables, a1 , a2 , . . . , an are their scalar
coefficients, and b is a constant. The Greek capital S, Σ, stands for
“Sum” and indicates that the terms of the equation, formed by iterat-
ing i from 1 to n, are to be summed. The expression is an abbrevia-
tion of
a1 x1 + a2 x2 + · · · + a n x n = b
Both the xi and the ai as well as b can take values from the set of
real numbers R. The most general notation for a system of m linear
,92 math syllabus
equations is:
n
∑ ai1 xi = b1
i =1
n
∑ ai2 xi = b2
i =1
..
.
n
∑ aim xi = bm
i =1
where aim is the coefficient for the i-th variable in the the m-th equa-
tion. These expressions are an abbreviation of
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
.. .. ..
. . .
am1 x1 + am2 x2 + · · · + amn xn = bm
Solution sets
In linear algebra we are interested in solutions to linear equations.
We often think of “solutions” as single values or a countable number
of values that the variables can assume to make both sides of an
equation equal. For example, x + 2 = 3 yields a single value for x,
1, that makes both sides of the equation equal. Or for the equation
x2 = 4 the two solutions x = 2 and x = −2 make both sides of the
equation equal. However, in algebra in general, and also in linear
algebra, we often encounter solutions that do not yield a single or
even a countable number of values for the variables, but solutions
that consist of uncountable14 (“infinite”) sets of values. In such cases, 14
“Uncountable” means that there
exists no one-to-one mapping with the
some or all variables can assume values in ranges of the set of real
(countable) set N of natural numbers.
numbers R, or over the whole set of real numbers. In linear algebra, Even though there are infinitely many
this situation is particularly interesting when there are multiple natural numbers, you can count them.
However, there is no way to count the
variables. We will demonstrate this principle using a few very simple set R, i.e. to map every number in R
examples. one-to-one with a number in N and vice
The simplest linear equation is versa.
ax = 0
and another, just a little more complicated equation is
ax = b
Let’s study the first equation, ax = 0. When you’re not too careful
you could think that x = 0 is “the solution” to that equation. How-
ever, what if a = 0? In that case, actually any x from the set R of real
, introduction 93
numbers will solve the equation! Therefore, whether the solution is a
single number (x = 0) or an uncountable set of numbers R15 depends 15
{ x | x ∈ R} is the technical way of
on the value of the scalar a. For the second equation a single solution writing this solution in “set builder”
notation. In words: the set of all
could be x = ba . However, this is only a valid solution if a 6= 0. In numbers x where x is an element of the
the case that a = 0 and b 6= 0 the equation has no solution, or in real numbers.
the language of set theory: the solution is the empty set ∅. However,
when a = 0 and b = 0, we get the same situation as with the first
equation, and any member x of the real numbers R will solve this
equation again. Concluding:
Remark The solution to the linear equation
ax = b
is
∅ when
a = 0 and b 6= 0,
x= R when a=0 and b = 0,
b
a when a 6= 0
Take-home message: a “solution” may be an empty set, a single number,
multiple (countably many) numbers, or a set of infinitely many numbers. In
other words, we generalize the concept of a solution to a set of objects that is
not necessarily countable.
These are rather trivial solution sets: they are a bit boring or they
don’t have much “intricacy” or “structure”. Then again, we were only
looking at the simplest possible linear equation. Things become a
little more interesting with this linear equation:
ax + by = c (27)
where x and y are variables and a, b and c are scalars. What can we
say about the solution of this equation? First of all, since the equation
has two variables, the solution will always look like a combination of
two numbers, one for x and a second for y. Such combinations of two
or more numbers are written between brackets as ( x, y, z, . . .), and are
called “tuples”. Tuples are like sets, only the elements have a specific
place, i.e. they are ordered lists of elements. You have to write the
numbers in a specific order to know which value belongs to which
variable. This is unlike ordinary sets, written with curly brackets like
{ x, y, z, . . .}, which behave like bags of objects16 . The order in which 16
Furthermore, duplicate occurrences
the objects are written is irrelevant there. of elements in ordinary sets have no
relevance, i.e. {2, 3, 3} = {2, 3}, which is
To solve eq. (27) we could do what you are used to do in algebra, certainly not true for tuples:(2, 3, 3) 6=
namely try to get one variable on the left side of the equation, let’s (2, 3).
say x:
ax = c − by therefore
c b
x= a − ay
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