Download Basis and Dimension The Dimension Theorem Every basis of V has

Previous lectures: Basis and Dimension
The Dimension Theorem
Every basis of V has the same size.
Theorem Suppose that V is a d–dimensional vector space
a) Every linearly independent subset of V can be
extended to a basis of V .
b) Every spanning set in V contains a basis.
Useful Fact Suppose V is a d–dimensional vector space
and that y1 , y2 , . . . , yd are vectors in V . Then the following are equivalent.
a) {y1 , y2 , . . . , yd } is a basis of V .
b) {y1 , y2 , . . . , yd } are linearly independent.
c) {y1 , y2 , . . . , yd } span V .
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This lecture: Lagrange Polynomials
There are ”natural” bases for Vector Spaces Rn and Pn .
In general, the best choice for the basis of a vector space
is often determined by the problem itself. Lagrange polynomials are an example of a special basis of Pn chosen
to solve a specific problem, namely curve fitting.
Lagrange polynomials are used to solve the problem of
finding a polynomial of minimal degree that passes through
a given set of points in the xy plane. Knowing this polynomial then allows predictions to be made about the y
values corresponding to other values of x. This process
is called interpolation if the x values lie in the range of
x used to define the Lagrange polynomials.
Suppose we consider the set of n+1 points
(c0 , k0 ), (c1 , k1 ), (c2 , k2 ), ......, (cn , kn )
in the xy plane. We assume that all the x values c0 , c1 , ....., cn
are different. We can display the points in tabulated form
as:
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x values
y values
c0
k0
c1
k1
c2
k2
...
...
cn
kn
Through two data points (c0 , k0 ) and (c1 , k1 ) we can fit
a straight line.
Through three data points (c0 , k0 ), (c1 , k1 ) and (c2 , k2 )
we can fit a parabola.
With n+1 data points (c0 , k0 ), (c1 , k1 ), (c2 , k2 ),.....(cn , kn ),
a polynomial of degree n is generally the one of minimum degree that fits the points exactly.
To find such a polynomial, we construct a special basis
of the vector space Pn making use of the x values from
the data table. This basis consists of n + 1 polynomials
of degree n,
(p0 , p1 , p2 , ....., pn ) satisfying the conditions
• pi (cj ) = 0 if i 6= j, and
• pi (ci ) = 1
for all i = 0, 1, 2, ....., n. That is, pi (x) is zero at all x
data values except ci and pi (x) is 1 when x = ci .
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Theorem If (p0 , p1 , p2 , ....., pn ) are polynomials in Pn
satisfying the conditions:
• pi (cj ) = 0 if i 6= j, and
• pi (ci ) = 1
for all i = 0, 1, 2, , ...., n, then (p0 , p1 , p2 , ....., pn ) is a
basis of Pn .
Proof First we prove linear independence.
Suppose that
a0 p0 (x) + a1 p1 (x) + a2 p2 (x) + ..... + an pn (x) = 0
for all values of x.
Substituting x = c0 gives
a0 × 1 + a1 × 0 + a2 × 0 + ..... + an × 0 = 0,
that is a0 = 0
Substituting x = c1 gives
a0 × 0 + a1 × 1 + a2 × 0 + ..... + an × 0 = 0,
that is a1 = 0
Continuing in this way, it is clear that all scalars
a0 , a1 , ....., an must be zero.
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This proves the linear independence of (p0 , p1 , p2 , ...., pn ).
Hence Span(p0 , p1 , p2 , ...., pn ) = Pn ,
Therefore (p0 , p1 , p2 , ...., pn ) is a basis of Pn .
Example Suppose the data is given by the Table:
x values
y values
0
1
1
3
2
5
From the data given, the Lagrange basis is (p0 ,p1 ,p2 ),
where
p0 (x) = (x − 1)(x − 2)/((0 − 1)(0 − 2))
= (x − 1)(x − 2)/2
p1 (x) = x(x − 2)/(1(1 − 2))
= −x(x − 2)
p2 (x) = x(x − 1)/(2(2 − 1))
= x(x − 1)/2
The polynomial that fits the data exactly is given by the
linear combination of the Lagrange basis vectors with the
y values as coefficients:
p(x) = 1p0 (x) + 3p1 (x) + 5p2 (x)
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= (x − 1)(x − 2)/2 − 3x(x − 2) + 5x(x − 1)/2
= 2x + 1
Observe the connection between Lagrange polynomials
and partial fractions. Suppose we had to break down the
expression
2x+1
x(x−1)(x−2)
into partial fractions.
The expression becomes
(1/2)(1/x) − 3(1/(x − 1)) + (5/2)(1/(x − 2))
For a different set of y values, such as:
x values
y values
0
1
1
3
2
7
the Lagrange polynomial is:
p(x) = 1p0 (x) + 3p1 (x) + 7p2 (x)
= (x − 1)(x − 2)/2 − 3x(x − 2) + 7x(x − 1)/2
= x2 + x + 1
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