## Wednesday, May 17, 2023

### Solving Simple Problems In Linear Algebra (1)

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, v2,  v3........v n be elements of V. Also let x 1, x 2, x 3...... x n  be numbers. Then it is possible to form an expression of the type:

x 1 v 1 + x 2  v 2 + x 3 v 3 +.............x n v n

which is called a linear combination of  v 1, v2v3........v n

The set of all linear combinations of v 1, v2v3........v n is a subspace of V,

Yet another example: let A be a vector in R 3. Let W be the set of all elements B in R 3 such that 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, v2v3........v n are linearly dependent over a field F if there exist elements in F not all equal to zero such that:

a 1 v 1 + a 2 v 2 + ..............a n v n = 0

If, on the other hand, there do not exist such numbers a1, a2 etc. we say that the elements v 1, v2v3........v n are linearly independent.

Now, if elements v 1, v2v3........v n of the vector space V generate V and also are linearly independent, then (v 1, v2v3........v n) is called a basis of V. One can also say that those elements v 1, v2v3........v n form a basis of V.

To fix ideas, let W f  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 W f.  As a further illustration, let V be a vector space and let (v 1, v2,  v3........v n) be a basis of 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 2 + x 3  v 3 +.............x 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)?

The coordinates are (3,5) with respect to the basis {exp(t), exp(2t)} .

Example Problem (1):

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

Solution

Let a, b be two numbers associated with some vector space - call it 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 R2.

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 R2, let (a,b) be an arbitrary
element of R2 and write out:

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)