EXAMPLES OF COORDINATE VECTORS

TERRY A. LORING

The book defines coordinate vector with respect to an ordered basis on page 157. As faras I can tell, the book has no examples of finding coordinate vectors, only eamples ofchanging coordinate systems.

Example 1. One ordered basis of vector space of two-by-two matrices is [E1, E2, E3, E4]where

E1 =

[

1 00 0

]

, E1 =

[

0 10 0

]

, E1 =

[

0 01 0

]

, E1 =

[

0 00 1

]

.

What is the coordinate vector of

A =

[

2 11 −1

]

w.r.t. this?

Remark 1. The solution is easy. We need to solve

x1

[

1 00 0

]

+ x2

[

0 10 0

]

+ x3

[

0 01 0

]

+ x4

[

0 00 1

]

=

[

2 11 −1

]

and that is so full of zeros we can skip writing out equations. The solution is

2

[

1 00 0

]

+ 1

[

0 10 0

]

+ 1

[

0 01 0

]

+ (−1)

[

0 00 1

]

=

[

2 11 −1

]

.

The coordinate vector is

211−1

Example 2. One ordered basis of vector space of two-by-two matrices is [E1, E2, E3, E4]where

E1 =

[

1 00 0

]

, E1 =

[

0 10 0

]

, E1 =

[

0 01 0

]

, E1 =

[

0 00 1

]

.

What is the matrix W that has coordinate vector

5432

w.r.t. this?

1

EXAMPLES OF COORDINATE VECTORS 2

Remark 2. The solution is easy. We need to solve

5

[

1 00 0

]

+ 4

[

0 10 0

]

+ 3

[

0 01 0

]

+ 2

[

0 00 1

]

= W

and that is just a little arithmetic:

W =

[

5 43 2

]

.

Example 3. Another ordered basis of vector space of two-by-two matrices is [F1, F2, F3, F4]where

F1 =

[

1 00 1

]

, F1 =

[

0 11 0

]

, F1 =

[

1 00 −1

]

, F1 =

[

0 1−1 0

]

.

(1) What is the coordinate vector of

A =

[

2 20 0

]

w.r.t. this?(2) What has the coordinate vector

5432

w.r.t. this ordered basis?Remark 3.

(1) The solution is a little harder. We need to solve

x1

[

1 00 1

]

+ x2

[

0 11 0

]

+ x3

[

1 00 −1

]

+ x4

[

0 1−1 0

]

=

[

2 20 0

]

and that works out as

x1 + x3 = 2x2 + x4 = 2x2 − x4 = 0

x1 − x3 = 0

EXAMPLES OF COORDINATE VECTORS 3

I can do this in my head, but to show the general proceedure we have augmentedmatrix, and a reduction, as follows:

1 0 1 0 20 1 0 1 20 1 0 −1 01 0 −1 0 0

∼

1 0 1 0 20 1 0 1 20 1 0 −1 00 0 −2 0 −2

∼

1 0 1 0 20 1 0 1 20 0 0 −2 −20 0 −2 0 −2

∼

1 0 1 0 20 1 0 1 20 0 0 1 10 0 1 0 1

∼

1 0 0 0 10 1 0 0 10 0 0 1 10 0 1 0 1

so the solution to the equations is

x1 = 1, x2 = 1, x3 = 1, x4 = 1

and the coordinate vector is

1111

.

(2)

5

[

1 00 1

]

+ 4

[

0 11 0

]

+ 3

[

1 00 −1

]

+ 2

[

0 1−1 0

]

=

[

8 62 2

]

so it is[

8 62 2

]

that has the specified coordinate vector.

Example 4. One ordered basis of P2 is [q1, q2, q3] where

q1(x) = x2− 2,

q2(x) = x2− 2x,

EXAMPLES OF COORDINATE VECTORS 4

q3(x) = x + 1.

What is the coordinate vector of p w.r.t. this if

p(x) = (x − 1)2?

Remark 4. We need to solve

r1

(

x2− 2

)

+ r2

(

x2− 2x

)

+ r3 (x + 1) = (x − 1)2

and sor1

(

x2− 2

)

+ r2

(

x2− 2x

)

+ r3 (x + 1) = x2− 2x + 1

and so(r1 + r2)x

2 + (−2r2 + r3)x + (−2r1 + r3) = 1x2− 2x + 1

and sor1 + r2 = 1

− 2r2 + r3 = −2−2r1 + r3 = 1

which has solution

r1 = −1

4r2 =

5

4r3 =

1

2.

Just to check:

−1

4

(

x2− 2

)

+5

4

(

x2− 2x

)

+1

2(x + 1)

= −1

4x2 +

1

2+

5

4x2

−5

2x +

1

2x +

1

2= x2

− 2x + 1

= (x2− 1)2.

Ok, so the answer is that the coordinate vector is

−1

45

41

2