Linear Algebra, Theory And Applications, 2012a

174 SPECTRAL THEORY
Theorem 7.4.4 Let A be an n × n matrix. Then there exists a unitary matrix U such that
U ∗ AU = T, (7.11)
where T is an upper triangular matrix having the eigenvalues of A on the main diagonal
listed according to multiplicity as roots of the characteristic equation.
Proof: The theorem is clearly true if A is a 1 × 1 matrix. Just let U =1the1× 1
matrix which has 1 down the main diagonal and zeros elsewhere. Suppose it is true for
(n − 1) × (n − 1) matrices and let A be an n × n matrix. Then let v 1 be a unit eigenvector
for A . Then there exists λ 1 such that
Av 1 = λ 1 v 1 , |v 1 | =1.
Extend {v 1 } to a basis and then use Lemma 7.4.1 to obtain {v 1 , ··· , v n }, an orthonormal
basis in F n .LetU 0 be a matrix whose i th column is v i . Then from the above, it follows U 0
is unitary. Then U0 ∗ AU 0 is of the form
⎛
⎜
⎝
λ 1 ∗ ··· ∗
0
. A 1
0
where A 1 is an n − 1 × n − 1 matrix. Now by induction there exists an (n − 1) × (n − 1)
unitary matrix Ũ1 such that
Ũ1 ∗ A 1 Ũ 1 = T n−1 ,
an upper triangular matrix. Consider
( ) 1 0
U 1 ≡
0 Ũ1
⎞
⎟
⎠
This is a unitary matrix and
( 1 0
U1 ∗ U0 ∗ AU 0 U 1 =
0 Ũ 1
∗
)(
λ1 ∗
)( 1 0
0 A 1 0 Ũ1
)
=
( )
λ1 ∗
≡ T
0 T n−1
where T is upper triangular. Then let U = U 0 U 1 . Since (U 0 U 1 ) ∗ = U1 ∗ U0 ∗ , it follows A
is similar to T and that U 0 U 1 is unitary. Hence A and T have the same characteristic
polynomials and since the eigenvalues of T are the diagonal entries listed according to
algebraic multiplicity,
As a simple consequence of the above theorem, here is an interesting lemma.
Lemma 7.4.5 Let A be of the form
⎛
⎞
P 1 ··· ∗
⎜
A =
.
⎝
.
. ..
. ⎟
. ⎠
0 ··· P s
where P k is an m k × m k matrix. Then
det (A) = ∏ k
det (P k ) .
Also, the eigenvalues of A consist of the union of the eigenvalues of the P j .