It was shown in Theorem 21.2.1 that A^{−1} exists if and only if det
(A )
≠0 when there is even a formula for
the inverse. Recall also that an eigenvector for λ is a nonzero vector x such that Ax = λx where λ is called
an eigenvalue. Thus you have
(A − λI)
x = 0 for x≠0. If
(A − λI)
^{−1} were to exist, then you could
multiply by it on the left and obtain x = 0 after all. Therefore, it must be the case that det
(A − λI)
= 0.
This yields a polynomial of degree n equal to 0. This polynomial is called the characteristic polynomial.For example, consider
That on the left equals a polynomial of degree 3 which when factored yields
(1− λ)(λ− 1)(λ− 2)
Therefore, the possible eigenvalues are 1,1,2. Note how the 1 is listed twice. This is because it occurs twice
as a root of the characteristic polynomial. Also, if M^{−1} does not exist where M is an n × n matrix, then
this means that the columns of M cannot be lineraly independent since if they were, then by Theorem 7.5.2M^{−1} would exist. Thus if A − λI fails to have an inverse as above, then the columns are not
independent and so there exists a nonzero x such that
(A − λI)
x = 0. Thus we have the following
proposition.
Proposition 21.2.5The eigenvalues of an n × n matrix are the roots of det
(A − λI)
= 0.Corresponding to each of these λ is an eigenvector.
Note that if A = S^{−1}BS, then A,B have the same characteristic polynomial, hence the same
eigenvalues. (They might have different eigenvectors and usually will.) To see this, note that from the
properties of determinants
( −1 −1 ) ( −1 )
det(A − λI) = det(S B)S − λS IS = det S ((B − λI))S
= det S−1 det(B − λI)det(S) = det S− 1S det(B − λI)
= det(I)det(B − λI) = det(B − λI) (21.4)
Definition 21.2.6Let A be n × n. Thentrace
(A)
≡∑_{i=1}^{n}A_{ii}.
Proposition 21.2.7Let A be m × n and let B be n × m. Thentrace
(AB )
= trace
(BA )
. Alsofor square matrices A,B, if A = S^{−1}BS, thentrace
As to the claim about the determinant, it follows from the properties of the determinant that
( ) ( )
det(A ) = detS −1BS = det S−1 det(B )det(S )
( −1 )
= det(B )det S S = det(B )■
These two, the trace and the determinant are two of the so called principal invariantsof a 3 × 3
matrix. The reason these are called invariants is that they are the same for A and B if these two
are related as described in the above proposition. In this case, the other principal invariant
is
1 2 1 ( 2)
2 (trace(A)) − 2 trace A
It turns out these are related to the coefficients of the characteristic polynomial defined as
det(A − λI)
and discussed below.
To see this last is also an invariant, the above proposition implies
The physical reason these are important is that their invariance implies they do not change when one uses
a different coordinate system to describe points. That which is physically meaningful cannot depend on
coordinate system because such coordinate systems are purely artificial constructions used
to identify points. Therefore, the principal invariants are good for formulating physical laws.
This is as far as we go here. To see much more on these ideas, you should take a course on
continuum mechanics. However, the trace and determinant also have a very interesting relation to
eigenvalues.
Theorem 21.2.8The trace of a matrix is the sum of its eigenvalues listed according to multiplicityas a root of the characteristic polynomial. Also, the determinant of the matrix equals the product ofits eigenvalues.
Proof:Let A be an n × n matrix. By Schur’s theorem, there is unitary U such that
U ∗AU = T
where T is upper triangular. The characteristic polynomial of T is
(λ − μ1)
(λ− μ2)
⋅⋅⋅
(λ− μn)
where
μ_{1},
⋅⋅⋅
,μ_{n} are the diagonal entries of T. From the above discussion 21.4, these must also be the eigenvalues
of A listed according to multiplicity since these two matrices A,T have the same characteristic polynomial.
By Proposition 21.2.7A,T have the same determinant, but since T is upper triangular, the product of its
diagonal entries is the product of the eigenvalues of A and this is the common value of the determinant of
these two matrices. ■
Example 21.2.9Find the eigenvalues of the following matrix.
( )
10 12 1
A = |( − 8 − 9 0 |)
7 8 0
You take det
(A − λI)
and after much fussing with details, you get the following for the characteristic
polynomial.
− X3 + X2 + X − 1
Thus the eigenvalues are the roots of this polynomial. These roots are 1,1,−1 when listed according
to multiplicity. You can use the above Theorem 21.2.8 as a way to check whether you likely
have this right. Indeed, when you add these together, you get 1. When you take the trace
of the above matrix, you get 1. This is a little reassurance that you didn’t make a mistake.
Note that the determinant of the above matrix is −1 which also equals the product of these
eigenvalues.