Consider the collection of n×n matrices consisting of complex numbers M_{n×n}. Now consider the special
matrices E_{ij} which has a 1 in the i^{th} row and j^{th} column and zeros in all other positions. Then according
to the way we add and multiply matrices by scalars, M_{n×n} can be considered as ℂ^{n2
}. For example,
consider the case where n = 2. Instead of writing
( )
1 2 3 4
, we bend it and write it as
( )
1 2
3 4
.
Thus, if a_{ij} is the entry in ij^{th} position of one of these matrices, then we can recover A by forming the
sum
∑ ∑
aijEij (7.1)
i k
(7.1)
Thus these E_{ij} span the n × n matrices which, as just noted, can be considered as vectors in ℂ^{n2
}.
Furthermore, these matrices E_{ij} are independent because if the above sum in 7.1 equals 0, then since a_{ij} is
the entry in the ij^{th} position, it follows that a_{ij} = 0. Thus these matrices are a basis for M_{n×n} and the
dimension of M_{n×n} = n^{2} which we already knew from the above identification of M_{n×n} with
ℂ^{n2
}.
Define A^{0}≡ I for any matrix A ∈ M_{n×n}. Consider the matrices
I,A,A2,⋅⋅⋅,An2
There are n^{2} + 1 of these matrices and so they can’t be independent. Hence there are scalars c_{0},
⋅⋅⋅
,c_{n2} not
all zero such that
2
c0I + c1A+ ⋅⋅⋅+ cn2An = 0
In other words, there is a polynomial
q(λ) = cmλm + ⋅⋅⋅+ c1λ + c0
such that
q(A) ≡ cmAm + ⋅⋅⋅+ c1A + c0I = 0 in Mn ×n
Out of all such polynomials, let
ˆp
(λ)
be one which has the smallest degree. Denote by p
(λ)
the polynomial
which results by dividing by the leading coefficient. Thus p
(λ)
is the monic polynomial of smallest degree
which has p
(A )
= 0.
Definition 7.1.1The minimum polynomial for an n × n matrix A is the polynomial which hassmallest degree and is monic such that p
(A)
= 0.In fact, it is unique and you might think aboutwhy this is so using the division algorithm.
Now we can give the definition of eigenvalues and eigenvectors.
Definition 7.1.2Let A be an n×n matrix. A NONZERO VECTOR x is said to be an eigenvector forA if there is some number λ such that
Ax = λx
This number is called an eigenvalue.
It turns out that every n × n matrix has an eigenvalue and that in fact every root of the
minimum polynomial is an eigenvalue. Recall that by the fundamental theorem of algebra, (See
Section 1.9), the minimum polynomial has a root. In fact, we can completely factor the minimum
polynomial.
Proposition 7.1.3Let p
(λ)
be a monic polynomial of degree m ≥ 1 having complex coefficients. Thenthere are complex numbers μ_{1},
⋅⋅⋅
,μ_{m}, possibly not all distince such that
p(λ) = (λ− μ )(λ− μ )⋅⋅⋅(λ− μ )
1 2 m
Proof:If m = 1, there is nothing to show. Suppose then that the Proposition is true for some m ≥ 1
and suppose p
(λ)
is a monic polynomial of degree m + 1.
From the fundamental theorem of algebra, there is a root to p
(λ)
. Denote this root as μ_{1}. From Lemma
1.8.2, the division algorithm,
p(λ) = (λ − μ1)k(λ)+ r(λ)
where the degree of r
(λ)
is less than 1 or else r
(λ)
= 0. Thus r
(λ)
= r. However, if we evaluate both sides
at λ = μ_{1} we get p
(μ1)
= 0 = r and so
p(λ) = (λ − μ1)k(λ)
Now k
(λ)
is also a monic polynomial which can be seen by comparing the leading coefficient of both sides
and it has degree m. Therefore, by induction, there are m complex numbers μ_{2},μ_{3},
The minimum polynomial can be computed although it might seem a little tedious. In the above
discussion, the minimum polynomial is only known to have degree no more than n^{2}. Actually it can be
shown that the degree of the minimum polynomial is never more than n although it might be less than n.
We will show this later as part of the theory of the determinant but in the meantime, one should go ahead
and use it. Here is an example.
Example 7.1.5Let
( )
A = 2 1
1 3
Find its minimum polynomial.
The matrices are I,
( )
2 1
1 3
,
( )
2 1
1 3
^{2}. These will end up being linearly dependent. They are
( )
1 0
0 1
,
( )
2 1
1 3
,
( )
5 5
5 10
. The polynomial is obtained by finding a linear combination of these
equal to 0. Lets make these into column vectors and use row operations.
( )
1 2 5
|| 0 1 5 ||
|( 0 1 5 |)
1 3 10
Now we row reduce this to get
( )
1 0 − 5
|| 0 1 5 ||
|( 0 0 0 |)
0 0 0
Thus, as explained earlier, the last column is −5 times the first added to 5 times the second.
Thus
( )2 ( ) ( )
2 1 = − 5 1 0 + 5 2 1
1 3 0 1 1 3
You can see from the row reduced echelon form that no smaller linear combination relating the matrices
I,A,A^{2} is possible. Hence the minimal polynomial is
λ2 − 5λ+ 5
The eigenvalues are therefore, the roots of this polynomial. They are
√ - √-
5 + 1 5, 5− 1 5
2 2 2 2
Now one can find eigenvectors associated with these. Consider the first of them. We want a
nonzero vector x such that
We can arrange them as column vectors in ℂ^{9} as done earlier, but it might be easier to simply look at the
entries in a single row or column. Lets pick the first column of each. Thus the augmented matrix to solve
would be
Thus column one of A^{2} equals −2 times column one of I added to three times column one of A.
Thus consider the polynomial λ^{2}− 3λ + 2. This seems to work in so far as the first column of
A is concerned and there is no polynomial of smaller degree which will work. Therefore, lets
check to see if this sends A to 0. If it does, then it must be the minimal polynomial. When you
do the computations, you find that this indeed does send A to 0 and so it is the minimum
polynomial. The eigenvalues are 1 and 2. You can now find the eigenvectors for these using row
operations.