lecture notes on the Cayley-Hamilton theorem

lecture notes on the Cayley-Hamilton theorem

1

Overview

You should all know about the

characteristic polynomial

of a

square
matrix

,

A

. To calculate the characteristic polynomial of

A

, one
subtracts a variable, say

t

, from the

diagonal

entries of

A

, and
then takes the

determinant

of the result. In other words, letting

p

A

⁢

(

t

)

denote the characteristic polynomial of

A

, one has

p

A

⁢

(

t

)

=

det

⁡

(

A

-

t

⁢

I

)

,

where

I

as usual denotes the

identity matrix

.
For example, set

A

=

[

1

2

3

0

-

2

1

1

1

0

]

.

Evaluating the determinant of

A

-

t

⁢

I

one gets

p

A

⁢

(

t

)

=

-

t

3

-

t

2

+

6

⁢

t

+

7

.

Now the interesting thing about square matrices is that one can do

algebra

with them. So if

A

is a

3

×

3

matrix then

A

2

,

A

3

, indeed every power of

A

will also be a

3

×

3

matrix.
Indeed, one can take any

polynomial

p

⁢

(

t

)

, and happily plug

A

into
it. The result will be some other

3

×

3

matrix. The obvious
question now is: what will happen when one plugs a square matrix

A

into its own characteristic polynomial? Let’s see what happens for
the sample

3

×

3

matrix above. Straightforward calculations show
that

A

2

=

[

4

1

5

1

5

-

2

1

0

4

]

,

A

3

=

[

9

11

13

-

1

-

10

8

5

6

3

]

,

Next adding the various powers of

A

with the coefficients of
characteristic polynomial (note that one uses the identity matrix in
place of the

constants

) one gets

-

A

3

-

A

2

6

⁢

A

7

⁢

I

-

[

9

11

13

-

1

-

10

8

5

6

3

]

-

[

4

1

5

1

5

-

2

1

0

4

]

+

6

⁢

[

1

2

3

0

-

2

1

1

1

0

]

+

7

⁢

[

1

0

0

0

1

0

0

0

1

]

=

[

0

0

0

0

0

0

0

0

0

]

Zero! One gets zero. This seemingly miraculous answer is not a
coincidence. Indeed one gets zero regardless of what matrix one
starts with. I encourage you to try this with a few examples of your own.

Theorem 1

(Cayley-Hamilton).

Let

A

be an

n

×

n

matrix, and let

p

A

⁢

(

t

)

=

det

⁡

(

A

-

t

⁢

I

)

be the
corresponding characteristic polynomial. Then,

p

A

⁢

(

A

)

=

0

.

The goal of these notes will be to explain and prove the above

theorem

. There are various hidden reasons that make the

Cayley-Hamilton theorem

work. It is the purpose of these notes to
bring these reasons into the open.

2

The Gist of the Matter.

Indeed there are two factors that make the Cayley-Hamilton theorem
such a striking and interesting result. Recall that if

U

is an

n

-dimensional

vector space

, and

𝐮

1

,

𝐮

2

,

𝐮

3

,

…

,

𝐮

n

+

1

, are any

n

+

1

vectors in

U

, then there will some kind of
a linear

relation

between the

𝐮

i

’s, i.e. for some choice of
scalars

a

1

,

a

2

,

…

,

a

n

+

1

one will have

a

1

⁢

𝐮

1

+

a

2

⁢

𝐮

2

+

a

3

⁢

𝐮

3

+

…

⁢

a

n

+

1

⁢

𝐮

n

+

1

=

0

.

Now the space of

3

×

3

matrices is

9

-dimensional. Therefore
for every

3

×

3

matrix

A

there must be a linear relationship
between the

10

different

matrix powers

A

9

,

A

8

,

A

7

,

A

6

,

A

5

,

A

4

,

A

3

,

A

2

,

A

1

=

A

,

A

0

=

I

.

The “miracle” of the Cayley-Hamilton theorem is twofold. First, a
linear relation arises already for the powers

A

3

,

A

2

,

A

1

,

A

0

.
Second, the coefficients for this linear relation are precisely the
coefficients of the characteristic polynomial of

A

.

Let’s put it another way. Look at the first

column vectors

of the
matrices

A

3

,

A

2

,

A

1

,

A

0

, i.e. the vectors

[

9

-

1

5

]

,

[

4

1

1

]

,

[

1

0

1

]

,

[

1

0

0

]

.

Now

ℝ

3

is a

3

-dimensional vector space, and so there
should be a linear relation between the above

4

vectors. Indeed
there is: the coefficients of the linear relation are

-

1

,

-

1

,

6

,

7

( i.e.

-

1

times the first vector, plus

-

1

times the second, plus

6

times the third, plus

7

times the fourth is equal to zero —
try it yourself!). What about the second column vectors of

A

3

,

A

2

,

A

1

,

A

0

? Now the vectors in question are

[

11

-

10

6

]

,

[

1

5

0

]

,

[

2

-

2

1

]

,

[

0

1

0

]

.

Again, we have here

4

vectors from a

3

-dimensional vectors
space, and therefore there should be a linear relation between the
vectors. However by some miracle the coefficients of the linear
relation for the second column vectors are the same as the
coefficients of the linear relation between the first column vectors,
namely

-

1

,

-

1

,

6

,

7

. Furthermore, these coefficients are precisely
the coefficients of the characteristic polynomial:

-

t

3

-

t

2

+

6

⁢

t

+

7

.
Needless to say the third column vectors are joined in a linear
relation with the same coefficients:

-

1

,

-

1

,

6

,

7

. Why is this
happening?

3

The Cyclic Basis

Let’s look again at the first column vectors of the matrices

A

0

,

A

1

,

A

2

(recall that

A

0

is just the identity matrix):

𝐮

0

=

[

1

0

0

]

,

𝐮

1

=

[

1

0

1

]

,

𝐮

2

=

[

4

1

1

]

,

and let’s take these

3

vectors as a new basis,

𝐁

. A basis
obtained in this fashion, i.e. by starting with a vector and
successively applying a matrix to it, is called a cyclic basis. What
will be the representation of the matrix

A

relative to this cyclic
basis? Now

𝐮

0

is just the first

elementary

vector,

𝐞

1

.
Furthermore, note that

𝐮

1

is nothing but

A

⁢

𝐮

0

, and that

𝐮

2

=

A

⁢

𝐮

1

. Now

A

⁢

𝐮

2

is the first column vector of

A

3

and
we already determined the linear relation between the first column
vectors of

A

3

,

…

⁢

A

0

. The bottom line is that

A

⁢

𝐮

2

=

-

(

𝐮

2

-

6

⁢

𝐮

1

-

7

⁢

𝐮

0

)

,

and consequently

A

will have
the following appearance relative to the basis

𝐁

:

[

A

]

𝐁

=

[

0

0

7

1

0

6

0

1

-

1

]

The

transition matrix

P

from

𝐁

to the

standard basis

𝐞

1

,

𝐞

2

,

𝐞

3

is given by

P

=

[

1

1

4

0

0

1

0

1

1

]

.

Of course

P

is relevant to our discussion precisely because

[

A

]

𝐁

=

P

-

1

⁢

A

⁢

P

.

Proposition 1

.

Let

A

be an

n

×

n

matrix,

P

a

non-singular

n

×

n

matrix, and set

B

=

P

-

1

⁢

A

⁢

P

. The matrices

A

and

B

have
the same characteristic polynomials.

Proof.

The characteristic polynomial of

B

is given as

det

⁡

(

B

-

t

⁢

I

)

=

det

⁡

(

P

-

1

⁢

A

⁢

P

-

t

⁢

I

)

=

det

⁡

(

P

-

1

⁢

(

A

-

t

⁢

I

)

⁢

P

)

.

Recall that the determinant of a

product

is the product of the
determinants, and that the determinant of an

inverse

is the inverse of
the determinant. Therefore

det

⁡

(

B

-

t

⁢

I

)

=

det

⁡

(

P

-

1

)

⁢

det

⁡

(

A

-

t

⁢

I

)

⁢

det

⁡

(

P

)

=

det

⁡

(

A

-

t

⁢

I

)

.

∎

In other words, according to the above theorem we should expect the
characteristic polynomial of

[

A

]

𝐁

to be equal to the
characteristic polynomial of

A

. Let’s check this using a co-factor
expansion.

|

-

t

0

7

1

-

t

6

0

1

-

1

-

t

|

=

-

t

⁢

(

t

2

+

t

-

6

)

+

7

⋅

1

=

-

t

3

-

t

2

+

6

⁢

t

+

7

Also note that the last column of

[

A

]

𝐁

contains all but one of
the coefficients of the characteristic polynomial. This too is not a
coincidence.

Proposition 2

.

Consider an

n

×

n

matrix

B

such that the

j

th

column vector for

j

=

1

,

2

,

…

⁢

n

-

1

is the basic vector

e

j

+

1

, while the last column of

B

is the vector

[

-

b

0

,

-

b

1

,

…

,

-

b

n

-

1

]

T

. In other words

B

has the following
form:

B

=

[

0

0

0

…

0

-

b

0

1

0

0

…

0

-

b

1

0

1

0

…

0

-

b

2

0

0

1

…

0

-

b

3

⋮

⋮

⋮

⋱

⋮

⋮

0

0

0

…

1

-

b

n

-

1

]

Then, the characteristic polynomial of

B

is given by

(

-

1

)

n

⁢

p

B

⁢

(

t

)

=

t

n

+

b

n

-

1

⁢

t

n

-

1

+

…

+

b

2

⁢

t

2

+

b

1

⁢

t

+

b

0

.

Proof.

We will calculate the determinant of

B

-

t

⁢

I

by doing a co-factor
expansion along the first row. Let

B

1

be the matrix obtained by
deleting the first row and the first column from

B

-

t

⁢

I

, and let

D

be the matrix obtained by deleting the first row and the last column
from

B

-

t

⁢

I

. Doing a co-factor expansion along the top row, it is
easy to see that

det

⁡

(

B

-

t

⁢

I

)

=

-

t

⁢

det

⁡

(

B

1

)

+

(

-

1

)

n

⁢

b

0

⁢

det

⁡

(

D

)

.

Now

D

is an

upper triangular matrix

with ones on the diagonal, and therefore

det

⁡

(

D

)

=

1

. The matrix

B

1

, on the other hand has the same

structure

as

B

-

t

⁢

I

, only it’s 1 size smaller. To that end let

B

2

be the matrix obtained by deleting the first two rows and
columns from

B

-

t

⁢

I

. By the same reasoning as above it’s easy to
see that

det

⁡

(

B

1

)

=

-

t

⁢

det

⁡

(

B

2

)

+

(

-

1

)

n

-

1

⁢

b

1

,

and therefore

det

⁡

(

B

-

t

⁢

I

)

=

(

-

1

)

n

⁢

b

0

-

t

⁢

(

(

-

1

)

n

-

1

⁢

b

1

-

t

⁢

det

⁡

(

B

2

)

)

.

Continuing inductively we see that for even

n

,
the determinant of

B

-

t

⁢

I

will have the form:

b

0

-

t

⁢

(

-

b

1

-

t

⁢

(

b

2

-

t

⁢

(

-

b

3

-

t

⁢

(

…

)

)

)

)

=

b

0

+

b

1

⁢

t

+

b

2

⁢

t

2

+

b

3

⁢

t

3

+

…

+

b

n

-

1

⁢

t

n

-

1

+

t

n

For odd

n

,

det

⁡

(

B

-

t

⁢

I

)

will be just like the

formula

above, but multiplied through by a

negative

sign.
∎

4

Putting it all together

Thanks to

Propositions

1

and

2

we are
now in a position to understand and to prove the
Cayley-Hamilton Theorem. Let

A

be an

n

×

n

matrix. Start by
setting

𝐮

0

=

𝐞

1

, and then create a

sequence

of vectors by
successively applying

A

, i.e.

𝐮

1

=

A

⁢

𝐮

0

,

𝐮

2

=

A

⁢

𝐮

1

, etc.
Notice that

𝐮

k

=

A

k

⁢

𝐮

0

; in other words,

𝐮

k

is the first
column of the matrix

A

k

.

Next, suppose that the

n

vectors

𝐮

0

,

𝐮

1

,

𝐮

2

,

…

,

𝐮

n

-

1

form a basis,

𝐁

, of

ℝ

n

(There are matrices

A

for which this doesn’t happen, but we’ll consider this possibility
later.) There will therefore exist scalars

b

0

,

b

1

,

…

⁢

b

n

-

1

such that

𝐮

n

+

b

n

-

1

⁢

𝐮

n

-

1

+

…

+

b

1

⁢

𝐮

1

+

b

0

⁢

𝐮

0

=

0

.

Now the representation of

A

relative to the cyclic basis

𝐁

will
have the form

[

A

]

𝐁

=

[

0

0

0

…

0

-

b

0

1

0

0

…

0

-

b

1

0

1

0

…

0

-

b

2

0

0

1

…

0

-

b

3

⋮

⋮

⋮

⋱

⋮

⋮

0

0

0

…

1

-

b

n

-

1

]

By Proposition

1

the characteristic polynomial of

[

A

]

𝐁

is equal to the characteristic polynomial of

A

.
Furthermore, by Proposition

2

the characteristic
polynomial of

[

A

]

𝐁

is equal to

±

(

t

n

+

b

n

-

1

⁢

t

n

-

1

+

…

⁢

b

1

⁢

t

+

b

0

)

.

Only one

conclusion

is possible:

b

0

,

b

1

,

…

,

b

n

-

1

must be precisely the
coefficients of the characteristic polynomial of

A

. Let us
summarize these findings.

Proposition 3

.

Let

A

be an

n

×

n

matrix, with characteristic polynomial

p

A

⁢

(

t

)

=

±

(

t

n

+

b

n

-

1

⁢

t

n

-

1

+

…

+

b

1

⁢

t

+

b

0

)

.

Fix a
number

k

between

1

and

n

, and let

u

j

be the

k

th

column of the matrix

A

j

. If the vectors

u

0

,

u

1

,

…

,

u

n

-

1

form a basis of

R

n

, then
the vectors

u

0

,

u

1

,

…

,

u

n

-

1

,

u

n

satisfy

the
linear relation:

𝐮

n

+

b

n

-

1

⁢

𝐮

n

-

1

+

…

+

b

1

⁢

𝐮

1

+

b

0

⁢

𝐮

0

=

0

.

5

A Complication

We are almost done with the proof of the Cayley-Hamilton
Theorem. First, however, we must deal with the possibility that the
square matrix

A

is such that the column vectors of

A

0

,

A

1

,

…

,

A

n

-

1

do not form a basis. Consider, for example

A

=

[

1

-

1

2

1

4

-

4

1

2

-

2

]

An easy calculation shows that the characteristic polynomial is given
by

p

A

⁢

(

t

)

=

t

3

-

3

⁢

t

2

+

t

+

2

.

Writing down the sequence of powers of

A

:

A

3

A

2

A

1

A

0

[

3

-

2

4

2

15

-

14

2

7

-

6

]

[

2

-

1

2

1

7

-

6

1

3

-

2

]

[

1

-

1

2

1

4

-

4

1

2

-

2

]

[

1

0

0

0

1

0

0

0

1

]

we notice that the first columns do, in fact, obey a linear relation
with the coefficients of the characteristic polynomial:

[

3

2

2

]

-

3

⁢

[

2

1

1

]

+

[

1

1

1

]

+

2

⁢

[

1

0

0

]

=

[

0

0

0

]

.

(1)

However these first column vectors

do not

form a basis of

ℝ

3

, and therefore Proposition

3

is not enough
to explain why these vectors obey the above linear relation.

In order to find an explanation, let us proceed as follows. Just as
before, start by setting

𝐮

0

=

𝐞

1

, and

𝐮

1

=

A

⁢

𝐮

0

. If we
take

𝐮

2

=

A

⁢

𝐮

1

, then

𝐁

=

(

𝐮

0

,

𝐮

1

,

𝐮

2

)

will not form a
basis, so instead, let us choose

𝐮

2

that is

linearly independent

from

𝐮

0

and

𝐮

1

, thereby ensuring that

𝐁

is a basis.
There are many, many possible such choices for

𝐮

2

. To keep the
discussion concrete, let us take

𝐮

2

=

𝐞

3

=

[

0

,

0

,

1

]

T

. Note that

A

⁢

𝐮

0

=

𝐮

1

A

⁢

𝐮

1

=

[

2

,

1

,

1

]

T

=

𝐮

0

+

𝐮

1

.

A

⁢

𝐮

2

=

[

2

,

-

4

,

2

]

T

=

6

⁢

𝐮

0

-

4

⁢

𝐮

1

+

2

⁢

𝐮

2

.

Therefore, representing

A

relative to the basis

𝐁

we obtain

[

A

]

𝐁

=

[

0

1

6

1

1

-

4

0

0

2

]

By Proposition

1

, we know that the characteristic
polynomial of

[

A

]

𝐁

is equal to the characteristic polynomial of

A

. However, we know much more.

Proposition 4

.

Let

B

be an

n

×

n

matrix of the form

B

=

[

B

1

B

2

𝟎

B

3

]

where

B

1

is a

k

×

k

matrix,

B

2

is a

k

×

(

n

-

k

)

matrix, and

B

3

is a

(

n

-

k

)

×

(

n

-

k

)

matrix. Then, the characteristic
polynomial of

B

is the product of the characteristic polynomials of

B

1

and

B

3

, i.e.

p

B

⁢

(

t

)

=

p

B

1

⁢

(

t

)

×

p

B

3

⁢

(

t

)

.

Proof.

Note that

B

-

t

⁢

I

=

[

B

1

-

t

⁢

I

1

B

2

𝟎

B

3

-

t

⁢

I

3

]

where

I

1

is the

k

×

k

identity matrix, and

I

3

is the

(

n

-

k

)

×

(

n

-

k

)

identity matrix. The Proposition now follows from
the fact that the determinant of a matrix whose shape is like

B

is
the determinant of the upper-left block times the determinant of the
lower-right block.
∎

Thanks to Proposition

4

we know that the
characteristic polynomial of

[

A

]

𝐁

is a product of the
characteristic polynomial of the

2

×

2

matrix

[

0

1

1

1

]

and the characteristic polynomial of the

1

×

1

matrix

[

2

]

. In
other words,

p

[

A

]

𝐁

⁢

(

t

)

=

(

t

2

-

t

-

1

)

⁢

(

t

-

2

)

.

Furthermore by Proposition

2

we know that

A

⁢

𝐮

1

,

𝐮

1

,

𝐮

0

, i.e. the
first column vectors of

A

2

,

A

1

,

A

0

, obey a linear relation with
the coefficients of the polynomial

t

2

-

t

-

1

:

A

⁢

𝐮

1

-

𝐮

1

-

𝐮

0

=

𝟎

,

[

2

1

1

]

-

[

1

1

1

]

-

[

1

0

0

]

=

[

0

0

0

]

.

(2)

Multiplying this relation through by

A

wed deduce that the first
column vectors of

A

3

,

A

2

,

A

1

obey the same linear relation:

[

3

2

2

]

-

[

2

1

1

]

-

[

1

1

1

]

=

[

0

0

0

]

.

(3)

Next think about what it means to multiply a polynomial such as

t

2

-

t

-

1

by another polynomial such as

t

-

2

. Indeed, one can
structure the

multiplication

by multiplying the first polynomial
through by

t

, then multiplying it through by

-

2

, and then adding
the two terms:

t

3

-

t

2

-

t

-

2

⁢

t

2

2

⁢

t

2

t

3

-

3

⁢

t

2

+

t

+

2

The bottom line is, of course, just the characteristic polynomial
of

A

, and the whole idea behind the above calculation is that

p

A

⁢

(

t

)

can be “formed out of” the polynomial

t

2

-

t

-

1

. This
shows that we can combine relations (

2

) and
(

3

) and produce in the end the desired relation
(

1

). All we have to do is take relation (

3

),
and add to it

-

2

times the relation (

2

). This explains
why the first column vectors of

A

3

,

A

2

,

A

1

,

A

0

obey a linear
relation whose coefficients come from the characteristic polynomial of

A

.

Proposition 5

.

Let

A

be an

n

×

n

matrix, with characteristic polynomial

p

A

⁢

(

t

)

=

±

(

t

n

+

b

n

-

1

⁢

t

n

-

1

+

…

+

b

1

⁢

t

+

b

0

)

.

Fix a
number

k

between

1

and

n

, and let

u

j

be the

k

th

column of the matrix

A

j

.
The vectors

u

0

,

u

1

,

…

,

u

n

-

1

,

u

n

satisfy the
linear relation:

𝐮

n

+

b

n

-

1

⁢

𝐮

n

-

1

+

…

+

b

1

⁢

𝐮

1

+

b

0

⁢

𝐮

0

=

0

,

even if the vectors

u

0

,

u

1

,

…

,

u

n

-

1

do not form a
basis of

R

n

.

Proof.

Suppose that there is a number

m

<

n

such that

𝐮

m

can be given
as a

linear combination

of

𝐮

0

,

𝐮

1

,

…

,

𝐮

m

-

1

; let’s
say

𝐮

m

+

c

m

-

1

⁢

𝐮

m

-

1

+

…

+

c

1

⁢

𝐮

1

+

c

0

⁢

𝐮

0

=

0

.

Choose vectors

𝐯

m

,

𝐯

m

+

1

,

…

,

𝐯

n

-

1

so that the
list

𝐁

=

(

𝐮

0

,

𝐮

1

,

…

,

𝐮

m

-

1

,

𝐯

m

,

𝐯

m

+

1

,

…

,

𝐯

n

-

1

)

forms a basis of

ℝ

n

. Relative to this basis,

A

will have
the form

A

=

[

A

1

A

2

0

A

3

]

,

where the upper-left block has the form

A

1

=

[

0

0

…

0

-

c

0

1

0

…

0

-

c

1

0

1

…

0

-

c

2

⋮

⋮

⋱

⋮

⋮

0

0

…

1

-

c

m

-

1

]

By Proposition

2

,

p

A

1

⁢

(

t

)

=

±

(

t

m

+

c

m

-

1

⁢

t

m

-

1

+

…

+

c

1

⁢

t

+

c

0

)

.
Let

±

(

t

n

-

m

+

d

n

-

m

-

1

⁢

t

m

-

1

+

…

⁢

d

1

⁢

t

+

d

0

)

denote the characteristic polynomial of

A

3

.
By Proposition

4

, the characteristic polynomial of

[

A

]

𝐁

(which is equal to the characteristic polynomial of

A

) is
the product

p

A

1

⁢

(

t

)

×

p

A

3

⁢

(

t

)

, and therefore

p

A

⁢

(

t

)

can
be obtained by taking linear combinations of

p

A

1

⁢

(

t

)

times
various powers of

t

:

d

0

×

t

m

+

c

m

-

1

⁢

t

m

-

1

+

…

+

c

2

⁢

t

2

+

c

1

⁢

t

+

c

0

d

1

×

t

m

+

1

+

c

m

-

1

⁢

t

m

+

c

m

-

2

⁢

t

m

-

1

+

…

+

c

1

⁢

t

2

+

c

0

⁢

t

…

…

…

1

×

t

n

+

c

m

-

1

⁢

t

n

-

1

+

…

+

c

1

⁢

t

n

-

m

+

1

+

c

0

⁢

t

n

-

m

t

n

+

b

n

-

1

⁢

t

n

-

1

+

b

n

-

2

⁢

t

n

-

2

+

…

+

b

2

⁢

t

2

+

b

1

⁢

t

+

b

0

Corresponding to the above polynomials are the relations

𝐮

m

+

c

m

-

1

⁢

𝐮

m

-

1

+

…

+

c

2

⁢

𝐮

2

+

c

1

⁢

𝐮

1

+

c

0

⁢

𝐮

0

=

0

𝐮

m

+

1

+

c

m

-

1

⁢

𝐮

m

+

c

m

-

2

⁢

𝐮

m

-

1

+

…

+

c

1

⁢

𝐮

2

+

c

0

⁢

𝐮

1

=

0

…

…

…

𝐮

n

+

c

m

-

1

⁢

𝐮

n

-

1

+

…

+

c

1

⁢

𝐮

n

-

m

+

1

+

c

0

⁢

𝐮

n

-

m

=

0

Adding these relations in the same way as the polynomials yields
the desired relation:

𝐮

n

+

b

n

-

1

⁢

𝐮

n

-

1

+

…

+

b

1

⁢

𝐮

1

+

b

0

⁢

𝐮

0

=

0

,

∎

Now we really are finished. Thanks to Propositions

3

and

5

we know that for every

k

=

1

,

…

,

n

, the

k

th

column vectors of the matrices

A

n

,

A

n

-

1

,

…

,

A

1

,

A

0

obey a linear relation with the coefficients of the characteristic
polynomial of

A

. Since this is true for every column of the above
matrices, it is certainly true for the full matrix — and that is the
precisely the conclusion of the Cayley-Hamilton theorem.

Title

lecture notes on the Cayley-Hamilton theorem

Canonical name

LectureNotesOnTheCayleyHamiltonTheorem

Date of creation

2013-03-22 16:50:59

Last modified on

2013-03-22 16:50:59

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

8

Author

rmilson (146)

Entry type

Topic

Classification

msc 15A18

Classification

msc 15A15

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