4 © 2012 Pearson Education, Inc. Vector Spaces 4.4 COORDINATE SYSTEMS

4

© 2012 Pearson Education, Inc.

Vector Spaces

4.4COORDINATE SYSTEMS

Slide 4.4- 2 © 2012 Pearson Education, Inc.

THE UNIQUE REPRESENTATION THEOREM

Theorem 7: Let be a basis for vector space V. Then for each x in V, there exists a unique set of scalars c1, …, cn such that

----(1)

1{b ,...,b }n

1 1x b ... bn nc c



Definition: Suppose is a basis for V and x is in V. The coordinates of x relative to the basis (or the -coordinate of x) are the weights c1, …, cn such that .

1{b ,...,b }n

1 1x b ... bn nc c



If c1, …, cn are the -coordinates of x, then the vector in

[x]

is the coordinate vector of x (relative to ), or the -coordinate vector of x.

The mapping is the coordinate mapping (determined by ).

1

n

c

c

n

x x


COORDINATES IN When a basis for is fixed, the -coordinate

vector of a specified x is easily found, as in the example below.

Example 1: Let , , , and

. Find the coordinate vector [x] of x relative to .

Solution: The -coordinate c1, c2 of x satisfy

nn

1

2b

1

2

1b

1

4x

5

1 2{b ,b }

1 2

2 1 4

1 1 5c c

b1 b2 x


COORDINATES IN

or

----(3)

This equation can be solved by row operations on an augmented matrix or by using the inverse of the matrix on the left.

In any case, the solution is , . Thus and

.

n

1

2

2 1 4

1 1 5

c

c

b1 b2 x

1 3c 2 2c 1 2x 3b 2b

1

2

3x

2B

c

c


COORDINATES IN See the following figure.

The matrix in (3) changes the -coordinates of a vector x into the standard coordinates for x.

An analogous change of coordinates can be carriedout in for a basis .

Let P

n

n 1{b ,...,b }n

1 2b b bn


COORDINATES IN Then the vector equation

is equivalent to ----(4)

P is called the change-of-coordinates matrix from to the standard basis in .

Left-multiplication by P transforms the coordinate vector [x] into x.

Since the columns of P form a basis for , P is invertible (by the Invertible Matrix Theorem).

n

1 1 2 2x b b ... bn nc c c

x xP B B

n

n


COORDINATES IN

Left-multiplication by converts x into its -coordinate vector:

The correspondence , produced by , is the coordinate mapping.

Since is an invertible matrix, the coordinate mapping is a one-to-one linear transformation from onto , by the Invertible Matrix Theorem.

n

1P B

1x xP B B

x x 1P B

1P B

nn


THE COORDINATE MAPPING

Theorem 8: Let be a basis for a vector space V. Then the coordinate mapping is a one-to-one linear transformation from V onto .

1{b ,...,b }n x x

n


THE COORDINATE MAPPING Every vector space calculation in V is accurately

reproduced in W, and vice versa. In particular, any real vector space with a basis of n

vectors is indistinguishable from .

Example 2: Let , , ,

and . Then is a basis for . Determine if x is in H, and if it is,

find the coordinate vector of x relative to .

n

1

3

v 6

2

2

1

v 0

1

3

x 12

7

1 2{v ,v }1 2Span{v ,v }H



Solution: If x is in H, then the following vector equation is consistent:

The scalars c1 and c2, if they exist, are the -coordinates of x.

1 2

3 1 3

6 0 12

2 1 7

c c



Using row operations, we obtain

.

Thus , and [x] .

3 1 3 1 0 2

6 0 12 0 1 3

2 1 7 0 0 0

1 2c 2 3c 2

3



The coordinate system on H determined by is shown in the following figure.


4


Vector Spaces

4.5THE DIMENSION OF A VECTOR SPACE


DIMENSION OF A VECTOR SPACE

Theorem 9: If a vector space V has a basis

, then any set in V containing more than n vectors must be linearly dependent.

Theorem 9 implies that if a vector space V has a basis

, then each linearly independent set in V has no more than n vectors.

1{b ,...,b }n

1{b ,...,b }n


DIMENSION OF A VECTOR SPACE Definition: If V is spanned by a finite set, then V is

said to be finite-dimensional, and the dimension of V, written as dim V, is the number of vectors in a basis for V. The dimension of the zero vector space {0} is defined to be zero. If V is not spanned by a finite set, then V is said to be infinite-dimensional.

Example 1: Find the dimension of the subspace3 6

5 4: , , , in

2

5

a b c

a dH a b c d

b c d

d


DIMENSION OF A VECTOR SPACE

H is the set of all linear combinations of the vectors

, , ,

Clearly, , v2 is not a multiple of v1, but v3 is a multiple of v2.

By the Spanning Set Theorem, we may discard v3 and still have a set that spans H.

1

1

5v

0

0

2

3

0v

1

0

3

6

0v

2

0

4

0

4v

1

5

1v 0


SUBSPACES OF A FINITE-DIMENSIONAL SPACE

Finally, v4 is not a linear combination of v1 and v2.

So {v1, v2, v4} is linearly independent and hence is a basis for H.

Thus dim .

Theorem 11: Let H be a subspace of a finite-dimensional vector space V. Any linearly independent set in H can be expanded, if necessary, to a basis for H. Also, H is finite-dimensional and

3H

dim dimH V


THE BASIS THEOREM

Theorem 12: Let V be a p-dimensional vector space,

. Any linearly independent set of exactly p elements in V is automatically a basis for V. Any set of exactly p elements that spans V is automatically a basis for V.

1p


DIMENSIONS OF NUL A AND COL A

Thus, the dimension of Nul A is the number of free variables in the equation , and the dimension of Col A is the number of pivot columns in A.

Example 2: Find the dimensions of the null space and the column space of

x 0A

3 6 1 1 7

1 2 2 3 1

2 4 5 8 4

A


DIMENSIONS OF NUL A AND COL A

Solution: Row reduce the augmented matrix to echelon form:

There are three free variable—x2, x4 and x5.

Hence the dimension of Nul A is 3. Also dim Col because A has two pivot columns.

0A

1 2 2 3 1 0

0 0 1 2 2 0

0 0 0 0 0 0

2A


4


Vector Spaces

4.6RANK


THE ROW SPACE

If A is an matrix, each row of A has n entries and thus can be identified with a vector in .

The set of all linear combinations of the row vectors is called the row space of A and is denoted by Row A.

Each row has n entries, so Row A is a subspace of

. Since the rows of A are identified with the columns

of AT, we could also write Col AT in place of Row A.

m nn

n


THE ROW SPACE

Theorem 13: If two matrices A and B are row equivalent, then their row spaces are the same. If B is in echelon form, the nonzero rows of B form a basis for the row space of A as well as for that of B.


THE ROW SPACE

Example 1: Find bases for the row space, the column space, and the null space of the matrix

Solution: To find bases for the row space and the column space, row reduce A to an echelon form:

2 5 8 0 17

1 3 5 1 5

3 11 19 7 1

1 7 13 5 3


THE ROW SPACE

By Theorem 13, the first three rows of B form a basis for the row space of A (as well as for the row space of B).

ThusBasis for Row

1 3 5 1 5

0 1 2 2 7

0 0 0 4 20

0 0 0 0 0

A B

:{(1,3, 5,1,5),(0,1, 2,2, 7),(0,0,0, 4,20)}A


THE ROW SPACE For the column space, observe from B that the pivots

are in columns 1, 2, and 4. Hence columns 1, 2, and 4 of A (not B) form a basis

for Col A:

Basis for Col

Notice that any echelon form of A provides (in its nonzero rows) a basis for Row A and also identifies the pivot columns of A for Col A.

2 5 0

1 3 1: , ,

3 11 7

1 7 5

A


THE ROW SPACE

However, for Nul A, we need the reduced echelon form.

Further row operations on B yield

1 0 1 0 1

0 1 2 0 3

0 0 0 1 5

0 0 0 0 0

A B C


THE ROW SPACE

The equation is equivalent to , that is,

So , , , with x3 and x5 free variables.

x 0A x 0C

1 3 5

2 3 5

4 5

0

2 3 0

5 0

x x x

x x x

x x

1 3 5x x x 2 3 52 3x x x 4 55x x


THE ROW SPACE The calculations show that

Basis for Nul

Observe that, unlike the basis for Col A, the bases for Row A and Nul A have no simple connection with the entries in A itself.

1 1

2 3

: ,1 0

0 5

0 1

A


THE RANK THEOREM Definition: The rank of A is the dimension of the

column space of A. Since Row A is the same as Col AT, the dimension of

the row space of A is the rank of AT. The dimension of the null space is sometimes called

the nullity of A. Theorem 14: The dimensions of the column space

and the row space of an matrix A are equal. This common dimension, the rank of A, also equals the number of pivot positions in A and satisfies the equation

m n

rank dim Nul A A n


THE RANK THEOREM

number of number of number of

pivot columns nonpivot columns columns


THE RANK THEOREM Example 2:

a. If A is a matrix with a two-dimensional null space, what is the rank of A?

b. Could a matrix have a two-dimensional null space?

Solution:

a. Since A has 9 columns, , and hence rank .

b. No. If a matrix, call it B, has a two-dimensional null space, it would have to have rank 7, by the Rank Theorem.

7 9

6 9

(rank ) 2 9A 7A

6 9


THE INVERTIBLE MATRIX THEOREM (CONTINUED)

But the columns of B are vectors in , and so the dimension of Col B cannot exceed 6; that is, rank B cannot exceed 6.

Theorem: Let A be an matrix. Then the following statements are each equivalent to the statement that A is an invertible matrix.

m. The columns of A form a basis of .

n. Col

o. Dim Col

p. rank

6

n n

nnA

A nA n


Documents

4 © 2012 Pearson Education, Inc. Vector Spaces 4.4 COORDINATE SYSTEMS