Theorem 19.4.1 says 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
linearly independent since if they were, then by Theorem 18.10.12M^{−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 19.4.4The eigenvalues of an n × n matrix are the rootsof

det(A − λI) = 0.

Correspondingto each of these λ is an eigenvector. Every n × n matrix for n ≥ 1 haseigenvectors and eigenvalues in ℂ^{n}.

Proof:It only remains to consider the last claim. This claim follows from the
fundamental theorem of algebra, Theorem 15.13.3. Indeed, the characteristic polynomial
is a polynomial of degree n. It has a zero λ_{1} by the fundamental theorem of calculus.
Thus

det(A − λI) = (z − λ1)p2(z)

where p_{2}

(z)

is a polynomial of degree n − 1. Now apply the fundamental theorem of
algebra to this one and continue this process untill you obtain an expression of the
form

n
det(A − λI) = (z − λ1)⋅⋅⋅(z − λn)(− 1)

then there are n eigenvalues with some maybe being repeated. ■

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

( ) ( )
det (A − λI) = det S−1BS − λS −1IS = det S −1(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 ) (19.2)