5Eigenvalues and eigenvectors

IA Vectors and Matrices

5.3 Transformation matrices

How do the components of a vector or a matrix change when we change the

basis?

Let

{e

1

, e

2

, ··· , e

n

}

and

{

˜

e

1

,

˜

e

2

, ··· ,

˜

e

n

}

be 2 different bases of

R

n

or

C

n

.

Then we can write

˜

e

j

=

n

X

i=1

P

ij

e

i

i.e.

P

ij

is the

i

th component of

˜

e

j

with respect to the basis

{e

1

, e

2

, ··· , e

n

}

.

Note that the sum is made as

P

ij

e

i

, not

P

ij

e

j

. This is different from the formula

for matrix multiplication.

Matrix

P

has as its columns the vectors

˜

e

j

relative to

{e

1

, e

2

, ··· , e

n

}

. So

P = (

˜

e

1

˜

e

2

···

˜

e

n

) and

P (e

i

) =

˜

e

i

Similarly, we can write

e

i

=

n

X

k=1

Q

ki

˜

e

k

with Q = (e

1

e

2

··· e

n

).

Substituting this into the equation for

˜

e

j

, we have

˜

e

j

=

n

X

i=1

n

X

k=1

Q

ki

˜

e

k

!

P

ij

=

n

X

k=1

˜

e

k

n

X

i=1

Q

ki

P

ij

!

But

˜

e

1

,

˜

e

2

, ··· ,

˜

e

n

are linearly independent, so this is only possible if

n

X

i=1

Q

ki

P

ij

= δ

kj

,

which is just a fancy way of saying QP = I, or Q = P

−1

.

5.3.1 Transformation law for vectors

With respect to basis

{e

i

}

,

u

=

P

n

i=1

u

i

e

i

. With respect to basis

{

˜

e

i

}

,

u

=

P

n

i=1

˜u

i

˜

e

i

. Note that this is the same vector

u

but has different components

with respect to different bases. Using the transformation matrix above for the

basis, we have

u =

n

X

j=1

˜u

j

n

X

i=1

P

ij

e

i

=

n

X

i=1

n

X

j=1

P

ij

˜u

j

e

i

By comparison, we know that

u

i

=

n

X

j=1

P

ij

˜u

j

Theorem.

Denote vector as

u

with respect to

{e

i

}

and

˜

u

with respect to

{

˜

e

i

}

.

Then

u = P ˜u and ˜u = P

−1

u

Example.

Take the first basis as

{e

1

= (1

,

0)

, e

2

= (0

,

1)

}

and the second as

{

˜

e

1

= (1, 1),

˜

e

2

= (−1, 1)}.

So

˜

e

1

= e

1

+ e

2

and

˜

e

2

= −e

1

+ e

2

. We have

P =

1 −1

1 1

.

Then for an arbitrary vector u, we have

u = u

1

e

1

+ u

2

e

2

= u

1

1

2

(

˜

e

1

−

˜

e

2

) + u

2

1

2

(

˜

e

1

+

˜

e

2

)

=

1

2

(u

1

+ u

2

)

˜

e

1

+

1

2

(−u

1

+ u

2

)

˜

e

2

.

Alternatively, using the formula above, we obtain

˜u = P

−1

u

=

1

2

1 1

−1 1

u

1

u

2

=

1

2

(u

1

+ u

2

)

1

2

(−u

1

+ u

2

)

Which agrees with the above direct expansion.

5.3.2 Transformation law for matrix

Consider a linear map α : C

n

→ C

n

with associated n × n matrix A. We have

u

0

= α(u) = Au.

Denote

u

and

u

0

as being with respect to basis

{e

i

}

(i.e. same basis in both

spaces), and ˜u, ˜u

0

with respect to {

˜

e

i

}.

Using what we’ve got above, we have

u

0

= Au

P ˜u

0

= AP

˜

u

˜u

0

= P

−1

AP ˜u

=

˜

A

˜

u

So

Theorem.

˜

A = P

−1

AP.

Example.

Consider the shear

S

λ

=

1 λ 0

0 1 0

0 0 1

with respect to the standard

basis. Choose a new set of basis vectors by rotating by θ about the e

3

axis:

˜

e

1

= cos θe

1

+ sin θe

2

˜

e

2

= −sin θe

1

+ cos θe

2

˜

e

3

= e

3

So we have

P =

cos θ −sin θ 0

sin θ cos θ 0

0 0 1

, P

−1

=

cos θ sin θ 0

−sin θ cos θ 0

0 0 1

Now use the basis transformation laws to obtain

˜

S

λ

=

1 + λ sin θ cos θ λ cos

2

θ 0

−λ sin

2

θ 1 − λ sin θ cos θ 0

0 0 1

Clearly this is much more complicated than our original basis. This shows that

choosing a sensible basis is important.

More generally, given

α

:

C

m

→ C

n

, given

x ∈ C

m

,

x

0

∈ C

n

with

x

0

=

Ax

.

We know that A is an n × m matrix.

Suppose

C

m

has a basis

{e

i

}

and

C

n

has a basis

{f

i

}

. Now change bases to

{

˜

e

i

} and {

˜

f

i

}.

We know that

x

=

P ˜x

with

P

being an

m × m

matrix, with

x

0

=

R

˜

x

0

with

R being an n × n matrix.

Combining both of these, we have

R

˜

x

0

= AP

˜

x

˜

x

0

= R

−1

AP ˜x

Therefore

˜

A = R

−1

AP .

Example.

Consider

α

:

R

3

→ R

2

, with respect to the standard bases in both

spaces,

A =

2 3 4

1 6 3

Use a new basis

2

1

,

1

5

in

R

2

and keep the standard basis in

R

3

. The basis

change matrix in R

3

is simply I, while

R =

2 1

1 5

, R

−1

=

1

9

5 −1

−1 2

is the transformation matrix for R

2

. So

˜

A =

2 1

1 5

2 3 4

1 6 3

I

=

1

9

5 −1

−1 2

2 3 4

1 6 3

=

1 1 17/9

0 1 2/9

We can alternatively do it this way: we know that

˜

f

1

=

2

1

,

˜

f

2

=

1

5

Then

we know that

˜

e

1

= e

1

7→ 2f

1

+ f

2

= f

1

˜

e

2

= e

2

7→ 3f

1

+ 6f

2

=

˜

f

1

+

˜

f

2

˜

e

3

= e

3

7→ 4f

1

+ 3f

2

=

17

9

˜

f

1

+

2

9

˜

f

2

and we can construct the matrix correspondingly.