CLASS NOTES FOR LINEAR MATH

1
CLASS NOTES FOR LINEAR MATH

• Section 3.4
• BASIS, COORDINATES AND DIMENSION

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DEFINITION OF A BASIS

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

3
EXAMPLES OF BASES

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

• A (the standard) basis for P2 is B 1, x, x2

PROOF Span B P2 because a linear combination
of the elements of B, i.e. ax2 bx c, is a
general element of P2. (Usually it takes more
work to show that a linear combination of the
elements of B span V.)
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EXAMPLES OF BASES

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

• A (the standard) basis for P2 is B 1, x, x2

PROOF Span B P2 because a linear combination
of the elements of B, i.e. ax2 bx c, is a
general element of P2. (Usually it takes more
work to show that a linear combination of the
elements of B span V.) Because two polynomials
are equal if their coefficients are equal, ax2
bx c 0 implies a 0, b 0 and c 0. This
means that the elements of B are linearly
independent.
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EXAMPLES OF BASES

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

• 2. The standard basis for R2 is B (1, 0),
  (0, 1)

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EXAMPLES OF BASES

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

• 2. A different basis for R2 is B
  (-1,2), (5,-9)

PROOF They are linearly independent, because
one is not a multiple of the other.
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EXAMPLES OF BASES

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

• 2. A different basis for R2 is B
  (-1,2), (5,-9)

PROOF They are linearly independent, because
one is not a multiple of the other. They span
because for a general element of R2, (a,b), we
can find constants m and n (in terms of the a and
b) so that m(-1,2) n(5,-9) (a,b).
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EXAMPLES OF BASES

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

• 2. A different basis for R2 is B
  (-1,2), (5,-9)

PROOF They are linearly independent, because
one is not a multiple of the other. They span
because for a general element of R2, (a,b), we
can find constants m and n (in terms of the a and
b) so that m(-1,2) n(5,-9) (a,b).
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EXAMPLES OF BASES

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

• 2. A different basis for R2 is B
  (-1,2), (5,-9)

PROOF They are linearly independent, because
one is not a multiple of the other. They span
because for a general element of R2, (a,b), we
can find constants m and n (in terms of the a and
b) so that m(-1,2) n(5,-9) (a,b).
(9a 5b)(-1,2) (2a b)(5,-9) (a,b).
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EXAMPLES OF BASES

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

• 3. Is (1,2,3), (4,5,6), (7,8,9) a basis for R3?

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EXAMPLES OF BASES

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

• 3. Is (1,2,3), (4,5,6), (7,8,9) a basis for R3?

Recall that the rows of a square matrix are
linearly independent if and only if the
determinant is not zero.
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EXAMPLES OF BASES

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

• 3. Is (1,2,3), (4,5,6), (7,8,9) a basis for R3?

Recall that the rows of a square matrix are
linearly independent if and only if the
determinant is not zero.
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EXAMPLES OF BASES

• A set B in a vector space V is called a basis for
  V if
• Span B V
• B is linearly independent

• 3. Is (1,2,3), (4,5,6), (7,8,9) a basis for R3?

Recall that the rows of a square matrix are
linearly independent if and only if the
determinant is not zero.
Since the second and third rows are multiples of
each other, the determinant is zero.
ANSWER NO
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COORDINATES
P(2, 4, 5)
The coordinates of a point P are 2, 4 and 5
because the standard basis is B (1, 0, 0), (0,
1, 0), (0, 0, 1) and P 2(1, 0, 0) 4(0, 1, 0)
5(0, 0, 1).
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COORDINATES
P(2, 4, 5)
The coordinates of a point P are 2, 4 and 5
because the standard basis is B (1, 0, 0), (0,
1, 0), (0, 0, 1) and P 2(1, 0, 0) 4(0, 1, 0)
5(0, 0, 1). We will usually write the results
of this as
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Discussion
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Discussion
For example the standard coordinates of P (3,4)
are 3 and 4, because 3(1,0) 4(0,1) (3,4)
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Discussion
For example the standard coordinates of P (3,4)
are 3 and 4, because 3(1,0) 4(0,1) (3,4)
but for B (-1,2), (5,-9), the B-coordinates
of (3,4) are 47 and 10 because 47(-1,2)
10(5,9) (3,4)
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Discussion
For example the standard coordinates of P (3,4)
are 3 and 4, because 3(1,0) 4(0,1) (3,4)
but for B (-1,2), (5,-9), the B-coordinates
of (3,4) are 47 and 10 because 47(-1,2)
10(5,9) (3,4) This came from example 2, where
we showed that (9a 5b)(-1,2) (2a b)(5,-9)
(a,b).
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PROPERTIES OF THE COORDINATE FUNCTION

• 1. Crda(ß) Crda(?) ? ß ? (Crd is
  one-to-one)

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PROPERTIES OF THE COORDINATE FUNCTION

• 1. Crda(ß) Crda(?) ? ß ? (Crd is
  one-to-one)
• 2. If X ?Mnx1 then there is ? ?V s.t. Crda(?) X
  (Crd is onto)

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PROPERTIES OF THE COORDINATE FUNCTION

• 1. Crda(ß) Crda(?) ? ß ? (Crd is
  one-to-one)
• 2. If X ?Mnx1 then there is ? ?V s.t. Crda(?) X
  (Crd is onto)
• 3. Crda(ß?) Crda(ß) Crda(?)
• 4. Crda(kß) k Crda(ß)

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PROPERTIES OF THE COORDINATE FUNCTION

• 1. Crda(ß) Crda(?) ? ß ? (Crd is
  one-to-one)
• 2. If X ?Mnx1 then there is ? ?V s.t. Crda(?) X
  (Crd is onto)
• 3. Crda(ß?) Crda(ß) Crda(?)
• 
  (Crd preserves operations)
• 4. Crda(kß) k Crda(ß)
• i.e. Crda V ?? Rn is an isomorphism (of vector
  spaces)

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EXAMPLE 1
Find if
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EXAMPLE 1
Find if
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EXAMPLE 1
Find if
So a 2, b 1, c 0 and d 1.
ANSWER
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EXAMPLE 2
Find Crda(x2) if a 1, 1 x, 1 x x2 for P3
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EXAMPLE 2
Find Crda(x2) if a 1, 1 x, 1 x x2 for P3
x2 a(1) b(1 x) c(1 x x2)
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EXAMPLE 2
Find Crda(x2) if a 1, 1 x, 1 x x2 for P3
x2 a(1) b(1 x) c(1 x x2)
x2 (abc) (b c)x cx2
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EXAMPLE 2
Find Crda(x2) if a 1, 1 x, 1 x x2 for P3
x2 a(1) b(1 x) c(1 x x2)
x2 (abc) (b c)x cx2
0 abc 0 b c 1 c
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EXAMPLE 2
Find Crda(x2) if a 1, 1 x, 1 x x2 for P3
x2 a(1) b(1 x) c(1 x x2)
x2 (abc) (b c)x cx2
0 abc 0 b c 1 c
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EXAMPLE 2
Find Crda(x2) if a 1, 1 x, 1 x x2 for P3
x2 a(1) b(1 x) c(1 x x2)
x2 (abc) (b c)x cx2
0 abc 0 b c 1 c
The inverse of the coefficient matrix is
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EXAMPLE 2
Find Crda(x2) if a 1, 1 x, 1 x x2 for P2
x2 a(1) b(1 x) c(1 x x2)
x2 (abc) (b c)x cx2
0 abc 0 b c 1 c
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EXAMPLE 2
Find Crda(x2) if a 1, 1 x, 1 x x2 for P2
x2 a(1) b(1 x) c(1 x x2)
x2 (abc) (b c)x cx2
0 abc 0 b c 1 c
ANSWER
So a 0, b -1 and c 1.
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EXAMPLE 3
Find Crda(x2) if a 1, x, x2 for P3
ANSWER ?
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Lemma 3.4
If V has a basis ai1n then any ß i1n1 is
linearly dependent in V.
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Theorem 3.5
If ai1n and ß i1m are both bases for V then n
m.
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DEFINITIONS

• The dimension of a vector space is the number of
  elements in any basis.

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DEFINITIONS

• The dimension of a vector space is the number of
  elements in any basis.
• A vector space is finite dimensional if the
  number of elements in a basis is finite.

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EXAMPLES

• The dimension of a vector space is the number of
  elements in any basis.

• 1. dim Pn n1
• 2. dim Rnxm nm

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DEFINITIONS

• A vector space is infinite dimensional if the
  number of elements in any basis is not finite.
• For example dim C1a,b ?

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End of Section 3.4