Page 189 - Matrices theory and applications

10. Approximation of Eigenvalues
172
Proposition 10.1.2 If M is tridiagonal Hermitian and if the entries
m j+1,j are nonzero (that is, if M is irreducible), then the eigenvalues of
M are real and simple. Furthermore, if M j is the (Hermitian, tridiagonal,
irreducible) matrix obtained by keeping only the j last rows and columns of
M, the eigenvalues of M j strictly separate those of M j+1 .
The separation, not necessarily strict, of the eigenvalues of M j+1 by those
of M j has already been proved, in a more general framework, in Theorem
3.3.3.
Proof
The eigenvalues of a Hermitian matrix are real. Since this matrix is
diagonalizable, Proposition 10.1.1 shows that the eigenvalues are simple.
Both properties can be deduced from the following analysis.
We proceed by induction on j.If j ≥ 1, we decompose the matrix M j+1
blockwise:
 
m ¯ a 0 ··· 0
a
 
 
0
 M j  ,

 .

 . . 

0
where a =0 and m ∈ IR, m> 0. Let P l be the characteristic polynomial
of M l . We compute that of M j+1 by expanding according to the elements
of the first column:
2
P j+1 (X)= mP j (X) −|a| P j−1 (X), (10.1)
where P 0 ≡ 1by convention.
The induction hypothesis is as follows: P j and P j−1 have real entries and
have respectively j and j − 1real roots µ 1 ,... ,µ j and σ 1 ,... ,σ j−1 ,with
µ 1 <σ 1 <µ 2 < ··· <σ j−1 <µ j .
In particular, they have no other roots, and their roots are simple. The signs
of the values of P j−1 at points µ j thus alternate. Since P j−1 is positive over
(σ j−1 , +∞), we have (−1) j−k P j−1 (µ k ) > 0.
This hypothesis clearly holds at step j =1. If j ≥ 2 and if it holds at
step j, then (10.1) shows that P j+1 ∈ IR[X]. Furthermore,
2
(−1) j−k P j+1 (µ k )= −|a| (−1) j−k P j−1 (µ k ) < 0.
From the intermediate value theorem, P j+1 possesses a root λ k in
(µ k−1 ,µ k ). Furthermore, P j+1 (µ j ) < 0, and P j+1 (x) is positive for x " 1;
hence there is also a root in (µ j , +∞). Likewise, P j+1 has a root in
(−∞,µ 1 ). Hence, P j+1 possesses j + 1 distinct real roots λ k ,with
λ 1 <µ 1 <λ 2 < ··· <µ j <λ j+1 .
Since P j+1 has degree j + 1, there is no root other than the λ k ’s, and these
are simple.