90 4. Particular Determinants
Whent 0 =1,tr=0,r>0,T 2 n=1,En=On,An= diag[1 0 0...0].
Hence,Pn=
1
2,Qn= 1, and the sign ofT 2 nis positive, which proves part(b) of the theorem.
The above theorem is applied in Section 6.10 on the Einstein and Ernstequations.
Exercise.Prove that
T
(n)
12=T
(n)
n− 1 ,n=T
(n+1)
1 n;1,n+14.5.3 Skew-Centrosymmetric Determinants........
The determinantAn=|aij|nis said to be skew-centrosymmetric if
an+1−i,n+1−j=−aij.InA 2 n+1, the element at the center, that is, in position (n+1,n+ 1), is
necessarily zero, but inA 2 n, no element is necessarily zero.
Exercises
1.Prove thatA 2 ncan be expressed as the product of two determinants ofordernwhich can be written in the form (P+Q)(P−Q) and hence asthe difference between two squares.2.Prove thatA 2 n+1 can be expressed as a determinant containing an(n+1)×(n+ 1) block of zero elements and is therefore zero.3.Prove that if the zero element at the center ofA 2 n+1is replaced byx,thenA 2 n+1can be expressed in the formx(p+q)(p−q).4.6 Hessenbergians
4.6.1 Definition and Recurrence Relation.........
The determinant
Hn=|aij|n,whereaij= 0 wheni−j>1 or whenj−i>1 is known as a Hessenberg
determinant or simply a Hessenbergian. Ifaij = 0 wheni−j>1, the
Hessenbergian takes the form
Hn=∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
a 11 a 12 a 13 ··· a 1 ,n− 1 a 1 na 21 a 22 a 23 ··· a 2 ,n− 1 a 2 na 32 a 33 ··· ··· ···a 43 ··· ··· ······ ··· ···an− 1 ,n− 1 an− 1 ,nan,n− 1 ann∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ n