Section 4.4 Coordinate Systems and 4.7 Change of Basis

Section 4.4 Coordinate systems and 4.7 Change of basis

College of William and Mary

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Pf. Since B spans V , there exist such scalars.

Suppose that x also has the representation x = d1b1 + d2b2 + ... + dnbn for scalars d1, d2,..., dn.

Then we have 0 = x − x = (c1 − d1)b1 + (c2 − d2)b2 + ... + (cn − dn)bn.

Since B is linearly independent, all the weights must be zeros. That is, c1 = d1, c2 = d2,..., cn = dn.

Unique Representation Theorem

Thm (Unique Representation Theorem) let B = {b1, b2,..., bn} be a basis for a vector space V . Then for each x ∈ V , there exists a unique set of scalars c1, c2,..., cn such that x = c1b1 + c2b2 + ... + cnbn.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Suppose that x also has the representation x = d1b1 + d2b2 + ... + dnbn for scalars d1, d2,..., dn.

Then we have 0 = x − x = (c1 − d1)b1 + (c2 − d2)b2 + ... + (cn − dn)bn.

Since B is linearly independent, all the weights must be zeros. That is, c1 = d1, c2 = d2,..., cn = dn.

Unique Representation Theorem

Pf. Since B spans V , there exist such scalars.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Then we have 0 = x − x = (c1 − d1)b1 + (c2 − d2)b2 + ... + (cn − dn)bn.

Since B is linearly independent, all the weights must be zeros. That is, c1 = d1, c2 = d2,..., cn = dn.

Unique Representation Theorem

Pf. Since B spans V , there exist such scalars.

Suppose that x also has the representation x = d1b1 + d2b2 + ... + dnbn for scalars d1, d2,..., dn.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Since B is linearly independent, all the weights must be zeros. That is, c1 = d1, c2 = d2,..., cn = dn.

Unique Representation Theorem

Pf. Since B spans V , there exist such scalars.

Suppose that x also has the representation x = d1b1 + d2b2 + ... + dnbn for scalars d1, d2,..., dn.

Then we have 0 = x − x = (c1 − d1)b1 + (c2 − d2)b2 + ... + (cn − dn)bn.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Unique Representation Theorem

Pf. Since B spans V , there exist such scalars.

Suppose that x also has the representation x = d1b1 + d2b2 + ... + dnbn for scalars d1, d2,..., dn.

Then we have 0 = x − x = (c1 − d1)b1 + (c2 − d2)b2 + ... + (cn − dn)bn.

Since B is linearly independent, all the weights must be zeros. That is, c1 = d1, c2 = d2,..., cn = dn.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis If c1, c2,..., cn are the B-coordinates of x, then the vector   c1  c2  [x]B =   is the coordinate vector of x, or the B-coordinate ... cn vector of x.

The mapping x → [x]B is the coordinate mapping determined by B. When a basis B is fixed for Rn, it is easy to find the B-coordinate vector of a specified x.

Coordinates

Deﬁnition: Suppose B = {b1, b2,..., bn} is a basis for V and x is in V . The coordinates of x relative to the basis B (or the B-coordinate of x) are the weights c1, c2,..., cn so that

c1b1 + c2b2 + ... + cnbn = x

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis The mapping x → [x]B is the coordinate mapping determined by B. When a basis B is fixed for Rn, it is easy to find the B-coordinate vector of a specified x.

Coordinates

Deﬁnition: Suppose B = {b1, b2,..., bn} is a basis for V and x is in V . The coordinates of x relative to the basis B (or the B-coordinate of x) are the weights c1, c2,..., cn so that

c1b1 + c2b2 + ... + cnbn = x

If c1, c2,..., cn are the B-coordinates of x, then the vector   c1  c2  [x]B =   is the coordinate vector of x, or the B-coordinate ... cn vector of x.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis When a basis B is fixed for Rn, it is easy to find the B-coordinate vector of a specified x.

Coordinates

Deﬁnition: Suppose B = {b1, b2,..., bn} is a basis for V and x is in V . The coordinates of x relative to the basis B (or the B-coordinate of x) are the weights c1, c2,..., cn so that

c1b1 + c2b2 + ... + cnbn = x

If c1, c2,..., cn are the B-coordinates of x, then the vector   c1  c2  [x]B =   is the coordinate vector of x, or the B-coordinate ... cn vector of x.

The mapping x → [x]B is the coordinate mapping determined by B.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Coordinates

Deﬁnition: Suppose B = {b1, b2,..., bn} is a basis for V and x is in V . The coordinates of x relative to the basis B (or the B-coordinate of x) are the weights c1, c2,..., cn so that

c1b1 + c2b2 + ... + cnbn = x

If c1, c2,..., cn are the B-coordinates of x, then the vector   c1  c2  [x]B =   is the coordinate vector of x, or the B-coordinate ... cn vector of x.

The mapping x → [x]B is the coordinate mapping determined by B. When a basis B is fixed for Rn, it is easy to find the B-coordinate vector of a specified x.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Sol. The B-coordinate c1, c2 of x satisfy c1b1 + c2b2 = x.

2 −1 c 4 So we have the following matrix equation 1 = . 1 1 c2 5

It can be easily solved (how many ways can you think of?) that c1 = 3 and c2 = 2.

3 So [x] = . B 2

Example

2 −1 4 Ex. Let b = , b = , x = and B = {b , b }. Find the 1 1 2 1 5 1 2 coordinate vector [x]B of x.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis 2 −1 c 4 So we have the following matrix equation 1 = . 1 1 c2 5

It can be easily solved (how many ways can you think of?) that c1 = 3 and c2 = 2.

3 So [x] = . B 2

Example

2 −1 4 Ex. Let b = , b = , x = and B = {b , b }. Find the 1 1 2 1 5 1 2 coordinate vector [x]B of x.

Sol. The B-coordinate c1, c2 of x satisfy c1b1 + c2b2 = x.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis It can be easily solved (how many ways can you think of?) that c1 = 3 and c2 = 2.

3 So [x] = . B 2

Example

2 −1 4 Ex. Let b = , b = , x = and B = {b , b }. Find the 1 1 2 1 5 1 2 coordinate vector [x]B of x.

Sol. The B-coordinate c1, c2 of x satisfy c1b1 + c2b2 = x.

2 −1 c 4 So we have the following matrix equation 1 = . 1 1 c2 5

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis 3 So [x] = . B 2

Example

2 −1 4 Ex. Let b = , b = , x = and B = {b , b }. Find the 1 1 2 1 5 1 2 coordinate vector [x]B of x.

Sol. The B-coordinate c1, c2 of x satisfy c1b1 + c2b2 = x.

2 −1 c 4 So we have the following matrix equation 1 = . 1 1 c2 5

It can be easily solved (how many ways can you think of?) that c1 = 3 and c2 = 2.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Example

2 −1 4 Ex. Let b = , b = , x = and B = {b , b }. Find the 1 1 2 1 5 1 2 coordinate vector [x]B of x.

Sol. The B-coordinate c1, c2 of x satisfy c1b1 + c2b2 = x.

2 −1 c 4 So we have the following matrix equation 1 = . 1 1 c2 5

It can be easily solved (how many ways can you think of?) that c1 = 3 and c2 = 2.

3 So [x] = . B 2

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis This approach can be generalized to Rn for a basis

B = {b1, b2,..., bn}

Let PB = [b1 b2 ... bn], which we call change-of-coordinates matrix from B to standard basis in Rn.

Then the vector equation x = c1b1 + c2b2 + ... + cnbn can be written as x = PB · [x]B

Coordinates in Rn

4 In the above example, we change the standard coordinate of x = 5 3 to the B-coordinate [x] = , through the matrix equation B 2

2 −1 x = [x] . 1 1 B

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Let PB = [b1 b2 ... bn], which we call change-of-coordinates matrix from B to standard basis in Rn.

Then the vector equation x = c1b1 + c2b2 + ... + cnbn can be written as x = PB · [x]B

Coordinates in Rn

4 In the above example, we change the standard coordinate of x = 5 3 to the B-coordinate [x] = , through the matrix equation B 2

2 −1 x = [x] . 1 1 B

This approach can be generalized to Rn for a basis

B = {b1, b2,..., bn}

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Then the vector equation x = c1b1 + c2b2 + ... + cnbn can be written as x = PB · [x]B

Coordinates in Rn

4 In the above example, we change the standard coordinate of x = 5 3 to the B-coordinate [x] = , through the matrix equation B 2

2 −1 x = [x] . 1 1 B

This approach can be generalized to Rn for a basis

B = {b1, b2,..., bn}

Let PB = [b1 b2 ... bn], which we call change-of-coordinates matrix from B to standard basis in Rn.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Coordinates in Rn

4 In the above example, we change the standard coordinate of x = 5 3 to the B-coordinate [x] = , through the matrix equation B 2

2 −1 x = [x] . 1 1 B

This approach can be generalized to Rn for a basis

B = {b1, b2,..., bn}

Let PB = [b1 b2 ... bn], which we call change-of-coordinates matrix from B to standard basis in Rn.

Then the vector equation x = c1b1 + c2b2 + ... + cnbn can be written as x = PB · [x]B

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis So we also have −1 [x]B = PB · x

−1 The mapping x → [x]B given by [x]B = PB · x is the coordinate mapping, which is one-to-one and onto, by IMT.

This is not only true from Rn to Rn, but also holds for a vector space with a basis of n vectors to Rn.

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

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis −1 The mapping x → [x]B given by [x]B = PB · x is the coordinate mapping, which is one-to-one and onto, by IMT.

This is not only true from Rn to Rn, but also holds for a vector space with a basis of n vectors to Rn.

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

So we also have −1 [x]B = PB · x

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis This is not only true from Rn to Rn, but also holds for a vector space with a basis of n vectors to Rn.

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

So we also have −1 [x]B = PB · x

−1 The mapping x → [x]B given by [x]B = PB · x is the coordinate mapping, which is one-to-one and onto, by IMT.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis n Since the columns of PB form a basis for R , PB is invertible (by the Invertible Matrix Theorem).

So we also have −1 [x]B = PB · x

−1 The mapping x → [x]B given by [x]B = PB · x is the coordinate mapping, which is one-to-one and onto, by IMT.

This is not only true from Rn to Rn, but also holds for a vector space with a basis of n vectors to Rn.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Pf. We need to show it is a linear transformation, one-to-one, and onto. To show it is a linear transformation, we need to check it preserves addition (of vectors) and scalar multiplication.

That is, [u + v]B = [u]B + [v]B and [cu]B = c[u]B .

Let u = c1b1 + c2b2 + ... + cnbn and v = d1b1 + d2b2 + ... + dnbn. Then

u + v = (c1 + d1)b1 + (c2 + d2)b2 + ... + (cn + dn)bn

and cu = (cc1)b1 + (cc2)b2 + ... + (ccn)bn

The Coordinate Mapping

Thm. Let B = {b1, b2,..., bn} be a basis for a vector space V . Then the coordinate mapping x → [x]B is a one-to-one and onto linear transformation from V to Rn.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis To show it is a linear transformation, we need to check it preserves addition (of vectors) and scalar multiplication.

That is, [u + v]B = [u]B + [v]B and [cu]B = c[u]B .

Let u = c1b1 + c2b2 + ... + cnbn and v = d1b1 + d2b2 + ... + dnbn. Then

u + v = (c1 + d1)b1 + (c2 + d2)b2 + ... + (cn + dn)bn

and cu = (cc1)b1 + (cc2)b2 + ... + (ccn)bn

The Coordinate Mapping

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis That is, [u + v]B = [u]B + [v]B and [cu]B = c[u]B .

Let u = c1b1 + c2b2 + ... + cnbn and v = d1b1 + d2b2 + ... + dnbn. Then

u + v = (c1 + d1)b1 + (c2 + d2)b2 + ... + (cn + dn)bn

and cu = (cc1)b1 + (cc2)b2 + ... + (ccn)bn

The Coordinate Mapping

Thm. Let B = {b1, b2,..., bn} be a basis for a vector space V . Then the coordinate mapping x → [x]B is a one-to-one and onto linear transformation from V to Rn. Pf. We need to show it is a linear transformation, one-to-one, and onto. To show it is a linear transformation, we need to check it preserves addition (of vectors) and scalar multiplication.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Let u = c1b1 + c2b2 + ... + cnbn and v = d1b1 + d2b2 + ... + dnbn. Then

u + v = (c1 + d1)b1 + (c2 + d2)b2 + ... + (cn + dn)bn

and cu = (cc1)b1 + (cc2)b2 + ... + (ccn)bn

The Coordinate Mapping

That is, [u + v]B = [u]B + [v]B and [cu]B = c[u]B .

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis The Coordinate Mapping

That is, [u + v]B = [u]B + [v]B and [cu]B = c[u]B .

Let u = c1b1 + c2b2 + ... + cnbn and v = d1b1 + d2b2 + ... + dnbn. Then

u + v = (c1 + d1)b1 + (c2 + d2)b2 + ... + (cn + dn)bn

and cu = (cc1)b1 + (cc2)b2 + ... + (ccn)bn

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis     cc1 c1 cc2  c2  And [cu]B =   = c   = c[u]B . ... ... ccn cn

We skip the proof for one-to-one and onto (homework).

      c1 + d1 c1 d1 c2 + d2  c2  d2  So [u + v]B =   =   +   = [u]B + [v]B .  ...  ... ... cn + dn cn dn

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis We skip the proof for one-to-one and onto (homework).

      c1 + d1 c1 d1 c2 + d2  c2  d2  So [u + v]B =   =   +   = [u]B + [v]B .  ...  ... ... cn + dn cn dn

    cc1 c1 cc2  c2  And [cu]B =   = c   = c[u]B . ... ... ccn cn

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis       c1 + d1 c1 d1 c2 + d2  c2  d2  So [u + v]B =   =   +   = [u]B + [v]B .  ...  ... ... cn + dn cn dn

    cc1 c1 cc2  c2  And [cu]B =   = c   = c[u]B . ... ... ccn cn

We skip the proof for one-to-one and onto (homework).

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis If u1,..., up are in V and if c1,..., cp are scalars, then

[c1u1 + c2u2 + ... + cpup]B = c1[u1]B + c2[u2]B + ... + cp[up]B In words, the above equation says that the B-coordinate vector of a linear combination of u1,..., up is the same linear combination of their coordinate vectors. The coordinate mapping in Theorem 8 is an important example of an isomorphism from V onto W . In general, a one-to-one linear transformation from a vector space V onto a vector space W is called an isomorphism from V onto W . The notation and terminology for V and W may diﬀer, but the two spaces are indistinguishable as vector spaces, if there is an isomorphism between them.

Isomorphism

The linearity of the coordinate mapping extends to linear combinations.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis In words, the above equation says that the B-coordinate vector of a linear combination of u1,..., up is the same linear combination of their coordinate vectors. The coordinate mapping in Theorem 8 is an important example of an isomorphism from V onto W . In general, a one-to-one linear transformation from a vector space V onto a vector space W is called an isomorphism from V onto W . The notation and terminology for V and W may diﬀer, but the two spaces are indistinguishable as vector spaces, if there is an isomorphism between them.

Isomorphism

The linearity of the coordinate mapping extends to linear combinations.

If u1,..., up are in V and if c1,..., cp are scalars, then

[c1u1 + c2u2 + ... + cpup]B = c1[u1]B + c2[u2]B + ... + cp[up]B

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis The coordinate mapping in Theorem 8 is an important example of an isomorphism from V onto W . In general, a one-to-one linear transformation from a vector space V onto a vector space W is called an isomorphism from V onto W . The notation and terminology for V and W may diﬀer, but the two spaces are indistinguishable as vector spaces, if there is an isomorphism between them.

Isomorphism

The linearity of the coordinate mapping extends to linear combinations.

If u1,..., up are in V and if c1,..., cp are scalars, then

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis In general, a one-to-one linear transformation from a vector space V onto a vector space W is called an isomorphism from V onto W . The notation and terminology for V and W may diﬀer, but the two spaces are indistinguishable as vector spaces, if there is an isomorphism between them.

Isomorphism

The linearity of the coordinate mapping extends to linear combinations.

If u1,..., up are in V and if c1,..., cp are scalars, then

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis The notation and terminology for V and W may diﬀer, but the two spaces are indistinguishable as vector spaces, if there is an isomorphism between them.

Isomorphism

The linearity of the coordinate mapping extends to linear combinations.

If u1,..., up are in V and if c1,..., cp are scalars, then

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Isomorphism

The linearity of the coordinate mapping extends to linear combinations.

If u1,..., up are in V and if c1,..., cp are scalars, then

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Pf. The coordinate vectors of the polynomials (over the standard basis B = {1, t, t2}) are (1, 0, 2), (4, 1, 5) and (3, 2, 0), respectively. Writing the vector as the columns of a matrix A, and determine their independence by row reducing the augmented matrix for Ax = 0:

1 4 3 0 1 4 3 0 0 1 2 0 → 0 1 2 0 2 5 0 0 0 0 0 0

The columns of A are linearly dependent, so the corresponding polynomials are linearly dependent.

Furthermore, c3 = 2c2 − 5c1. So we have

3 + 2t = 2(4 + t + 5t2) − 5(1 + 2t2).

Example

Ex. Use coordinate vectors to verify that the polynomials 2 2 1 + 2t , 4 + t + 5t and 3 + 2t are linearly dependent in P2.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Writing the vector as the columns of a matrix A, and determine their independence by row reducing the augmented matrix for Ax = 0:

1 4 3 0 1 4 3 0 0 1 2 0 → 0 1 2 0 2 5 0 0 0 0 0 0

The columns of A are linearly dependent, so the corresponding polynomials are linearly dependent.

Furthermore, c3 = 2c2 − 5c1. So we have

3 + 2t = 2(4 + t + 5t2) − 5(1 + 2t2).

Example

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis The columns of A are linearly dependent, so the corresponding polynomials are linearly dependent.

Furthermore, c3 = 2c2 − 5c1. So we have

3 + 2t = 2(4 + t + 5t2) − 5(1 + 2t2).

Example

Ex. Use coordinate vectors to verify that the polynomials 2 2 1 + 2t , 4 + t + 5t and 3 + 2t are linearly dependent in P2. Pf. The coordinate vectors of the polynomials (over the standard basis B = {1, t, t2}) are (1, 0, 2), (4, 1, 5) and (3, 2, 0), respectively. Writing the vector as the columns of a matrix A, and determine their independence by row reducing the augmented matrix for Ax = 0:

1 4 3 0 1 4 3 0 0 1 2 0 → 0 1 2 0 2 5 0 0 0 0 0 0

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Furthermore, c3 = 2c2 − 5c1. So we have

3 + 2t = 2(4 + t + 5t2) − 5(1 + 2t2).

Example

1 4 3 0 1 4 3 0 0 1 2 0 → 0 1 2 0 2 5 0 0 0 0 0 0

The columns of A are linearly dependent, so the corresponding polynomials are linearly dependent.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Example

1 4 3 0 1 4 3 0 0 1 2 0 → 0 1 2 0 2 5 0 0 0 0 0 0

The columns of A are linearly dependent, so the corresponding polynomials are linearly dependent.

Furthermore, c3 = 2c2 − 5c1. So we have

3 + 2t = 2(4 + t + 5t2) − 5(1 + 2t2).

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis T Namely, if x = x1b1 + x2b2 + ... + xnbn, then [x]B = [x1 x2 ... xn] .

In applications, we may know the B-coordinate for the vector, but need to know its C-coordinate for another basis C.

We will give a way to build connection between [x]B and [x]C , in particular, we will ﬁnd the change-of-coordinates matrix from B to C.

Basis and coordinates

When a basis B is given for an n-dimensional vector space V , every vector x in V can be uniquely identiﬁed by its B-coordinate vector [x]B .

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis In applications, we may know the B-coordinate for the vector, but need to know its C-coordinate for another basis C.

We will give a way to build connection between [x]B and [x]C , in particular, we will ﬁnd the change-of-coordinates matrix from B to C.

Basis and coordinates

When a basis B is given for an n-dimensional vector space V , every vector x in V can be uniquely identiﬁed by its B-coordinate vector [x]B .

T Namely, if x = x1b1 + x2b2 + ... + xnbn, then [x]B = [x1 x2 ... xn] .

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis We will give a way to build connection between [x]B and [x]C , in particular, we will ﬁnd the change-of-coordinates matrix from B to C.

Basis and coordinates

When a basis B is given for an n-dimensional vector space V , every vector x in V can be uniquely identiﬁed by its B-coordinate vector [x]B .

T Namely, if x = x1b1 + x2b2 + ... + xnbn, then [x]B = [x1 x2 ... xn] .

In applications, we may know the B-coordinate for the vector, but need to know its C-coordinate for another basis C.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Basis and coordinates

When a basis B is given for an n-dimensional vector space V , every vector x in V can be uniquely identiﬁed by its B-coordinate vector [x]B .

T Namely, if x = x1b1 + x2b2 + ... + xnbn, then [x]B = [x1 x2 ... xn] .

In applications, we may know the B-coordinate for the vector, but need to know its C-coordinate for another basis C.

We will give a way to build connection between [x]B and [x]C , in particular, we will ﬁnd the change-of-coordinates matrix from B to C.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis 3 Sol. Note that [x] = [3b + b ] = 3[b ] + [b ] = [b ] [b ] . C 1 2 C 1 C 2 C 1 C 2 C 1

T T As [b1]C = [4 1] , [b2]C = [−6 1] , we have

4 −6 3 6 [x] = = . C 1 1 1 4

Example

Ex. Consider two bases B = {b1, b2} and C = {c1, c2} for a vector space V , such that b1 = 4c1 + c2, b2 = −6c1 + c2

Suppose that x = 3b1 + b2, ﬁnd [x]C .

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis T T As [b1]C = [4 1] , [b2]C = [−6 1] , we have

4 −6 3 6 [x] = = . C 1 1 1 4

Example

Ex. Consider two bases B = {b1, b2} and C = {c1, c2} for a vector space V , such that b1 = 4c1 + c2, b2 = −6c1 + c2

Suppose that x = 3b1 + b2, ﬁnd [x]C .

3 Sol. Note that [x] = [3b + b ] = 3[b ] + [b ] = [b ] [b ] . C 1 2 C 1 C 2 C 1 C 2 C 1

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Example

Ex. Consider two bases B = {b1, b2} and C = {c1, c2} for a vector space V , such that b1 = 4c1 + c2, b2 = −6c1 + c2

Suppose that x = 3b1 + b2, ﬁnd [x]C .

3 Sol. Note that [x] = [3b + b ] = 3[b ] + [b ] = [b ] [b ] . C 1 2 C 1 C 2 C 1 C 2 C 1

T T As [b1]C = [4 1] , [b2]C = [−6 1] , we have

4 −6 3 6 [x] = = . C 1 1 1 4

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis THM. Let B = {b1, b2,..., bn} and C = {c1, c2,..., cn} be bases of a vector of a vector space V . Then there is a unique n × n matrix PB→C such that [x]C = PB→C · [x]B . Here the change-of-coordinates matrix from B to C is PB→C = [b1]C [b2]C ... [bn]C

Note that the columns of PB→C are linearly independent and it is a square matrix, it must be an invertible matrix. So we also have the following −1 [x]B = PB→C · [x]C . In other words, −1 PC→B = PB→C

Change of basis

The previous example actually can be generalized.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Note that the columns of PB→C are linearly independent and it is a square matrix, it must be an invertible matrix. So we also have the following −1 [x]B = PB→C · [x]C . In other words, −1 PC→B = PB→C

Change of basis

The previous example actually can be generalized.

THM. Let B = {b1, b2,..., bn} and C = {c1, c2,..., cn} be bases of a vector of a vector space V . Then there is a unique n × n matrix PB→C such that [x]C = PB→C · [x]B . Here the change-of-coordinates matrix from B to C is PB→C = [b1]C [b2]C ... [bn]C

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis In other words, −1 PC→B = PB→C

Change of basis

The previous example actually can be generalized.

Note that the columns of PB→C are linearly independent and it is a square matrix, it must be an invertible matrix. So we also have the following −1 [x]B = PB→C · [x]C .

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Change of basis

The previous example actually can be generalized.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis n Let B = {b1, b2,..., bn} be a basis of R and E = {e1, e2,..., en} be the standard basis of Rn.

Then [b1]E = b1. So PB→E = [ b1 b2 ... bn ] = PB .

n Let x = [x1 x2 ... xn] be a vector in R (by using the standard basis).

Then we have [x] = PB · [x]B n Let C be another basis of R . Then [x] = PC · [x]C = PB · [x]B .

That is, PB · [x]B = PC · [x]C

Change of basis in Rn

We now consider a special case, when the vector space is Rn.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Then [b1]E = b1. So PB→E = [ b1 b2 ... bn ] = PB .

n Let x = [x1 x2 ... xn] be a vector in R (by using the standard basis).

Then we have [x] = PB · [x]B n Let C be another basis of R . Then [x] = PC · [x]C = PB · [x]B .

That is, PB · [x]B = PC · [x]C

Change of basis in Rn

We now consider a special case, when the vector space is Rn.

n Let B = {b1, b2,..., bn} be a basis of R and E = {e1, e2,..., en} be the standard basis of Rn.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis n Let x = [x1 x2 ... xn] be a vector in R (by using the standard basis).

Then we have [x] = PB · [x]B n Let C be another basis of R . Then [x] = PC · [x]C = PB · [x]B .

That is, PB · [x]B = PC · [x]C

Change of basis in Rn

We now consider a special case, when the vector space is Rn.

n Let B = {b1, b2,..., bn} be a basis of R and E = {e1, e2,..., en} be the standard basis of Rn.

Then [b1]E = b1. So PB→E = [ b1 b2 ... bn ] = PB .

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Then we have [x] = PB · [x]B n Let C be another basis of R . Then [x] = PC · [x]C = PB · [x]B .

That is, PB · [x]B = PC · [x]C

Change of basis in Rn

We now consider a special case, when the vector space is Rn.

n Let B = {b1, b2,..., bn} be a basis of R and E = {e1, e2,..., en} be the standard basis of Rn.

Then [b1]E = b1. So PB→E = [ b1 b2 ... bn ] = PB .

n Let x = [x1 x2 ... xn] be a vector in R (by using the standard basis).

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis n Let C be another basis of R . Then [x] = PC · [x]C = PB · [x]B .

That is, PB · [x]B = PC · [x]C

Change of basis in Rn

We now consider a special case, when the vector space is Rn.

n Let B = {b1, b2,..., bn} be a basis of R and E = {e1, e2,..., en} be the standard basis of Rn.

Then [b1]E = b1. So PB→E = [ b1 b2 ... bn ] = PB .

n Let x = [x1 x2 ... xn] be a vector in R (by using the standard basis).

Then we have [x] = PB · [x]B

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis That is, PB · [x]B = PC · [x]C

Change of basis in Rn

We now consider a special case, when the vector space is Rn.

n Let B = {b1, b2,..., bn} be a basis of R and E = {e1, e2,..., en} be the standard basis of Rn.

Then [b1]E = b1. So PB→E = [ b1 b2 ... bn ] = PB .

n Let x = [x1 x2 ... xn] be a vector in R (by using the standard basis).

Then we have [x] = PB · [x]B n Let C be another basis of R . Then [x] = PC · [x]C = PB · [x]B .

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Change of basis in Rn

We now consider a special case, when the vector space is Rn.

n Let B = {b1, b2,..., bn} be a basis of R and E = {e1, e2,..., en} be the standard basis of Rn.

Then [b1]E = b1. So PB→E = [ b1 b2 ... bn ] = PB .

n Let x = [x1 x2 ... xn] be a vector in R (by using the standard basis).

Then we have [x] = PB · [x]B n Let C be another basis of R . Then [x] = PC · [x]C = PB · [x]B .

That is, PB · [x]B = PC · [x]C

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis I Write down the matrix [PC PB ] I Row reduce the matrix so that PC becomes In. I the matrix PB becomes PB→C .

−1 This is the change-of-coordinate matrix PB→C = PC · PB .

Note that

−1 −1 −1 PC · [PC PB ] = [PC · PC PC PB ] = [IPB→C ]

−1 So we actually have an algorithm to ﬁnd PB→C , instead of PC · PB :

So we have −1 [x]C = PC · PB · [x]B

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis I Write down the matrix [PC PB ] I Row reduce the matrix so that PC becomes In. I the matrix PB becomes PB→C .

Note that

−1 −1 −1 PC · [PC PB ] = [PC · PC PC PB ] = [IPB→C ]

−1 So we actually have an algorithm to ﬁnd PB→C , instead of PC · PB :

So we have −1 [x]C = PC · PB · [x]B −1 This is the change-of-coordinate matrix PB→C = PC · PB .

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis I Write down the matrix [PC PB ] I Row reduce the matrix so that PC becomes In. I the matrix PB becomes PB→C .

−1 So we actually have an algorithm to ﬁnd PB→C , instead of PC · PB :

So we have −1 [x]C = PC · PB · [x]B −1 This is the change-of-coordinate matrix PB→C = PC · PB .

Note that

−1 −1 −1 PC · [PC PB ] = [PC · PC PC PB ] = [IPB→C ]

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis I Write down the matrix [PC PB ] I Row reduce the matrix so that PC becomes In. I the matrix PB becomes PB→C .

So we have −1 [x]C = PC · PB · [x]B −1 This is the change-of-coordinate matrix PB→C = PC · PB .

Note that

−1 −1 −1 PC · [PC PB ] = [PC · PC PC PB ] = [IPB→C ]

−1 So we actually have an algorithm to ﬁnd PB→C , instead of PC · PB :

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis I Row reduce the matrix so that PC becomes In. I the matrix PB becomes PB→C .

So we have −1 [x]C = PC · PB · [x]B −1 This is the change-of-coordinate matrix PB→C = PC · PB .

Note that

−1 −1 −1 PC · [PC PB ] = [PC · PC PC PB ] = [IPB→C ]

−1 So we actually have an algorithm to ﬁnd PB→C , instead of PC · PB :

I Write down the matrix [PC PB ]

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis I the matrix PB becomes PB→C .

So we have −1 [x]C = PC · PB · [x]B −1 This is the change-of-coordinate matrix PB→C = PC · PB .

Note that

−1 −1 −1 PC · [PC PB ] = [PC · PC PC PB ] = [IPB→C ]

−1 So we actually have an algorithm to ﬁnd PB→C , instead of PC · PB :

I Write down the matrix [PC PB ] I Row reduce the matrix so that PC becomes In.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis So we have −1 [x]C = PC · PB · [x]B −1 This is the change-of-coordinate matrix PB→C = PC · PB .

Note that

−1 −1 −1 PC · [PC PB ] = [PC · PC PC PB ] = [IPB→C ]

−1 So we actually have an algorithm to ﬁnd PB→C , instead of PC · PB :

I Write down the matrix [PC PB ] I Row reduce the matrix so that PC becomes In. I the matrix PB becomes PB→C .

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis I Find the change-of-coordinate matrix from C to B. I Find the change-of-coordinate matrix from B to C.

To ﬁnd PC→B , we consider the matrix [PC PB ]:

1 −2 −7 −5 1 0 5 3 [P P ] = → B C −3 4 9 7 0 1 6 4

5 3 So P = . C→B 6 4 −1 To ﬁnd PB→C , we just need to ﬁnd PC→B :

1 4 −3 2 −3/2 P = P−1 = · = B→C C→B 2 −6 5 −3 5/2

Example

T T T T Ex. Let b1 = [1 − 3] , b2 = [−2 4] , c1 = [−7 9] , c2 = [−5 7] , and 2 consider the bases for R given by B = {b1, b2} and C = {c1, c2}.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis I Find the change-of-coordinate matrix from B to C.

To ﬁnd PC→B , we consider the matrix [PC PB ]:

1 −2 −7 −5 1 0 5 3 [P P ] = → B C −3 4 9 7 0 1 6 4

5 3 So P = . C→B 6 4 −1 To ﬁnd PB→C , we just need to ﬁnd PC→B :

1 4 −3 2 −3/2 P = P−1 = · = B→C C→B 2 −6 5 −3 5/2

Example

T T T T Ex. Let b1 = [1 − 3] , b2 = [−2 4] , c1 = [−7 9] , c2 = [−5 7] , and 2 consider the bases for R given by B = {b1, b2} and C = {c1, c2}.

I Find the change-of-coordinate matrix from C to B.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis To ﬁnd PC→B , we consider the matrix [PC PB ]:

1 −2 −7 −5 1 0 5 3 [P P ] = → B C −3 4 9 7 0 1 6 4

5 3 So P = . C→B 6 4 −1 To ﬁnd PB→C , we just need to ﬁnd PC→B :

1 4 −3 2 −3/2 P = P−1 = · = B→C C→B 2 −6 5 −3 5/2

Example

T T T T Ex. Let b1 = [1 − 3] , b2 = [−2 4] , c1 = [−7 9] , c2 = [−5 7] , and 2 consider the bases for R given by B = {b1, b2} and C = {c1, c2}.

I Find the change-of-coordinate matrix from C to B. I Find the change-of-coordinate matrix from B to C.

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis 5 3 So P = . C→B 6 4 −1 To ﬁnd PB→C , we just need to ﬁnd PC→B :

1 4 −3 2 −3/2 P = P−1 = · = B→C C→B 2 −6 5 −3 5/2

Example

T T T T Ex. Let b1 = [1 − 3] , b2 = [−2 4] , c1 = [−7 9] , c2 = [−5 7] , and 2 consider the bases for R given by B = {b1, b2} and C = {c1, c2}.

I Find the change-of-coordinate matrix from C to B. I Find the change-of-coordinate matrix from B to C.

To ﬁnd PC→B , we consider the matrix [PC PB ]:

1 −2 −7 −5 1 0 5 3 [P P ] = → B C −3 4 9 7 0 1 6 4

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis −1 To ﬁnd PB→C , we just need to ﬁnd PC→B :

1 4 −3 2 −3/2 P = P−1 = · = B→C C→B 2 −6 5 −3 5/2

Example

T T T T Ex. Let b1 = [1 − 3] , b2 = [−2 4] , c1 = [−7 9] , c2 = [−5 7] , and 2 consider the bases for R given by B = {b1, b2} and C = {c1, c2}.

I Find the change-of-coordinate matrix from C to B. I Find the change-of-coordinate matrix from B to C.

To ﬁnd PC→B , we consider the matrix [PC PB ]:

1 −2 −7 −5 1 0 5 3 [P P ] = → B C −3 4 9 7 0 1 6 4

5 3 So P = . C→B 6 4

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis Example

T T T T Ex. Let b1 = [1 − 3] , b2 = [−2 4] , c1 = [−7 9] , c2 = [−5 7] , and 2 consider the bases for R given by B = {b1, b2} and C = {c1, c2}.

I Find the change-of-coordinate matrix from C to B. I Find the change-of-coordinate matrix from B to C.

To ﬁnd PC→B , we consider the matrix [PC PB ]:

1 −2 −7 −5 1 0 5 3 [P P ] = → B C −3 4 9 7 0 1 6 4

5 3 So P = . C→B 6 4 −1 To ﬁnd PB→C , we just need to ﬁnd PC→B :

1 4 −3 2 −3/2 P = P−1 = · = B→C C→B 2 −6 5 −3 5/2

Gexin Yu [email protected] Section 4.4 Coordinate systems and 4.7 Change of basis