__ __* Preliminaries*:

The fundamental starting point for linear algebra is the concept of the **vector space. **Let **V** be an arbitrary vector
space and let **v _{1}**,

**v**,

_{2}**v**........

_{3}**v**be elements of

_{n}**V**. Also let x

**, x**

_{ 1}**, x**

_{ 2}**...... x**

_{ 3}**be numbers. Then it is possible to form an expression of the type:**

_{ n}x

_{ 1}**v**+ x

_{1}

_{ 2}**v**+ x

_{2}

_{ 3}**v**+.............x

_{3}

_{n}**v**

_{n}which is called a

*linear combination*of

**v**,

_{1}**v**,

_{2}**v**........

_{3}**v**.

_{n}**v**,

_{1}**v**,

_{2}**v**........

_{3}**v**is a

_{n}*subspace*of

**V**,

Yet another example: let

**A**be a vector in

**R**

**. Let**

^{ 3}**W**be the set of all elements B in

**R**

**such that**

^{ 3}**B**

**·**

**A**= 0, i.e. such that

**B**is perpendicular to

**A**. Then

**W**is a subspace of

**R**

**.**

^{ 3}An additional important consideration is whether elements of a vector space are

**linearly dependent**or linearly independent. We say the elements

**v**,

_{1}**v**,

_{2}**v**........

_{3}**v**are linearly

_{n}*dependent*over a field

**F**if there exist elements in

**F**not all equal to zero such that:

a

_{ 1}**v**+ a

_{1}

_{ 2}**v**+ ..............a

_{2}

_{n}**v**= 0

_{n}If, on the other hand, there

*do not exist*such numbers a1, a2 etc. we say that the elements

**v**,

_{1}**v**,

_{2}**v**........

_{3}**v**are

_{n}**linearly independent**.

Now, if elements

**v**,

_{1}**v**,

_{2}**v**........

_{3}**v**of the vector space

_{n}**V**generate

**V**and also are linearly independent, then (

**v**,

_{1}**v**,

_{2}**v**........

_{3}**v**) is called a

_{n}**basis**of

**V**. One can also say that those elements

**v**,

_{1}**v**,

_{2}**v**........

_{3}**v**form a basis of

_{n}**V**.

**W**be a vector space of functions generated by the two functions: exp(t) and exp(2t), then {exp(t), exp(2t)} is a basis of

_{f }**W**As a further illustration, let

_{f. }**V**be a vector space and let (

**v**,

_{1}**v**,

_{2}**v**........

_{3}**v**) be a basis of

_{n}**V**. The elements of

**V**can be represented by n-tuples relative to this basis, e.g. if an element

**v**of

**V**is written as a linear combination:

**v** = x_{ 1}**v _{1}** + x

_{ 2}**v**+ x

_{2}

_{ 3 }**v**+.............x

_{3}

_{n}**v**

_{n}Let **V **be the vector space of
functions generated by the two functions: exp(t) and exp(2t), then what are the
coordinates for f(**V**) = 3 exp(t) + 5
exp(2t)?

**Example Problem (1):**

Show that the vectors (1, 1) and (-3, 2) are linearly independent.

*Solution*:

**W**- such that:

a(1,1) + b(-3,2) = 0

Writing out the components as linear combinations:

a - 3b = 0 and a + 2b = 0

Then solve simultaneously:

a - 3b = 0

a + 2b = 0

----------

0 -5b = 0

or b = 0, so a = 0

Both a and b are equal to zero so the vectors are linearly independent.

*Example Problem (2):*Find the coordinates of (1, 0) with respect to the two vectors (1,1) and (-1, 2)

*Solution*:

We must find numbers a and b which meet the condition:

a(1, 1) + b(-1, 2) = (1, 0)

This can be rewritten:

a - b = 1 and a + 2b = 0

Solve simultaneously, by subtracting the 2nd from the 1st:

a - b = 1

a + 2b = 0

-----------

0 - 3b = 1

and 3b = -1, so b = - 1/3, then a = 1 + b = 1 - 1/3 = 2/3

Then the coordinates of (1, 0) with respect to (1, 1) and (-1,2) are: (2/3, -1/3)

**:**

*Example*Problem (3)Show that the vectors (1, 1) and (-1, 2) form a basis of

**R**

**.**

^{2}*Solution*: This requires showing; a) the vectors are linearly independent, and b) they generate

**R**

**.**

^{2}As before (earlier problems), we set out the condition via expression for linear independence:

a(1, 1) + b(-1, 2) = (0, 0)

-> a - b = 0 and a + 2b = 0

solve simultaneously by subtracting the 2nd from the 1st:

a - b = 0

a + 2b = 0

----------

0 - 3b = 0 so that b = 0 and a = 0

Thus the vectors are linearly independent.

(b) To show generation of

**R**

**, let (a,b) be an arbitrary element of**

^{2}**R**

**and write out:**

^{2}x (1, 1) + y(-1, 2) = (a, b)

which leads to the pair of simultaneous equations:

x - y = a and x + 2y = b

As before, subtracting the 2nd from the 1st eqn.

x - y = a

x + 2y = b

----------

0 - 3y = a - b or y = (b - a)/ 3

Therefore, (x,y) are the coordinates of (a,b) with respect to the basis {(1,1), (-1,2)}.

__Suggested Problems:__

1) Show the following vectors are linearly independent:

a) (π, 0) and (0, 1)

b) (1, 1, 0), (1, 1, 1) and (0, 1, -1)

2) Express X as a linear combination of the given vectors A, B and find the
coordinates of X with respect to A, B:

a) X = (1, 0), A = (1, 1), B = (0, 1)

b) X = (1,1), A = (2, 1), B = (-1, 0)

3)
Show the following vectors are linearly independent over **C** or **R**:

a) (1, 1, 1) and (0, 1, -2)

b) (-1, 1, 0) and (0, 1, 2)

c) (0, 1, 1), (0, 2, 1) and (1, 5, 3)

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