80 2 Algebra
The spectral mapping theorem.LetAbe ann×nmatrix with not necessarily distinct
eigenvaluesλ 1 ,λ 2 ,...,λn, and letP(x)be a polynomial. Then the eigenvalues of the
matrixP (A)areP(λ 1 ), P (λ 2 ),...,P(λn).
Proof.To prove this result we will apply a widely used idea (see for example the splitting
principle in algebraic topology). We will first assume that the eigenvalues ofAare all
distinct. ThenAcan be diagonalized by eigenvectors as
⎛
⎜⎜
⎜
⎝λ 1 0 ··· 0
0 λ 2 ··· 0
..
...
.
... ..
.
00 ···λn⎞
⎟⎟
⎟
⎠
,
and in the basis formed by the eigenvectors ofA, the matrixP (A)assumes the form
⎛
⎜
⎜⎜
⎝P(λ 1 ) 0 ··· 0
0 P(λ 2 )··· 0
..
...
.
... ..
.
00 ···P(λn)⎞
⎟
⎟⎟
⎠
.
The conclusion is now straightforward. In general, the characteristic polynomial of a
matrix depends continuously on the entries. Problem 172 in Section 2.2.4 proved that the
roots of a polynomial depend continuously on the coefficients. Hence the eigenvalues of
a matrix depend continuously on the entries.
The set of matrices with distinct eigenvalues is dense in the set of all matrices. To
prove this claim we need the notion of the discriminant of a polynomial. By definition, if
the zeros of a polynomial arex 1 ,x 2 ,...,xn, the discriminant is
∏
i<j(xi−xj)(^2). It is equal
to zero if and only if the polynomial has multiple zeros. Being a symmetric polynomial in
thexi’s, the discriminant is a polynomial in the coefficients. Therefore, the condition that
the eigenvalues of a matrix be not all distinct can be expressed as a polynomial equation
in the entries. By slightly varying the entries, we can violate this condition. Therefore,
arbitrarily close to any matrix there are matrices with distinct eigenvalues.
The conclusion of the spectral mapping theorem for an arbitrary matrix now follows
by a limiting argument.
We continue with two more elementary problems.
Example.LetA:V→WandB:W→Vbe linear maps between finite-dimensional
vector spaces. Prove that the linear mapsABandBAhave the same set of nonzero
eigenvalues, counted with multiplicities.