Determinants and Their Applications in Mathematical Physics

4.6 Hessenbergians 91

Ifaij= 0 whenj−i>1, the triangular array of zero elements appears

in the top right-hand corner.Hncan be expressed neatly in column vector

notation.

Let

Cjr=

[

a 1 ja 2 ja 3 j...arjOn−r

]T

n

, (4.6.2)

whereOirepresents an unbroken sequence ofizero elements. Then

Hn=

∣

∣

C 12 C 23 C 34 ...Cn− 1 ,nCnn

∣

∣

n

. (4.6.3)

Hessenbergians satisfy a simple recurrence relation.

Theorem 4.20.

Hn=(−1)

n− 1

n− 1

∑

r=0

(−1)

r

pr+1,nHr,H 0 =1,

where

pij=

{

aijaj,j− 1 aj− 1 ,j− 2 ···ai+2,i+1ai+1,i,j>i

aii,j=i.

Proof. ExpandingHnby the two nonzero elements in the last row,

Hn=annHn− 1 −an,n− 1 Kn− 1 ,

whereKn− 1 is a determinant of order (n−1) whose last row also contains

two nonzero elements. ExpandingKn− 1 in a similar manner,

Kn− 1 =an− 1 ,nHn− 2 −an− 1 ,n− 2 Kn− 2 ,

whereKn− 2 is a determinant of order (n−2) whose last row also contains

two nonzero elements. The theorem appears after these expansions are

repeated a sufficient number of times.

Illustration.

H 5 =

∣

∣C

12 C 23 C 34 C 45 C 55

∣

∣=a

55 H 4 −a 54

∣

∣C

12 C 23 C 34 C 54

∣

∣,

∣

∣C

12 C 23 C 34 C 54

∣

∣=a

45 H 3 −a 43

∣

∣C

12 C 23 C 53

∣

∣,

∣

∣C

12 C 23 C 53

∣

∣=a

35 H 2 −a 32

∣

∣C

12 C 52

∣

∣,

∣

∣C

12 C 52

∣

∣=a

25 H 1 −a 21 a 15 H 0.

Hence,

H 5 =a 55 H 4 −(a 45 a 54 )H 3 +(a 35 a 54 a 43 )H 2

−(a 25 a 54 a 43 a 32 )H 1 +(a 15 a 54 a 43 a 32 a 21 )H 0

=p 55 H 4 −p 45 H 3 +p 35 H 2 −p 25 H 1 +p 15 H 0.

Muir and Metzler use the termrecurrentwithout giving a definition of the

term. Arecurrentis any determinant which satisfies a recurrence relation.