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, v2, v3........v n.
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, v2, v3........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, v2, v3........v n are linearly independent.
Now, if elements v 1, v2, v3........v n of the vector space V generate V and also are linearly independent, then (v 1, v2, v3........v n) is called a basis of V. One can also say that those elements v 1, v2, v3........v n form a basis of V.
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:
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)
No comments:
Post a Comment