Page 304 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes A n  A 1k
( ) det   , 1 k . n
A
k
A  A
k 1 kk
Lemma: Let A be an invertible n × n matrix with entries in a field F. The following two statements
are equivalent:
(a) There is an upper triangular matrix P with P = 1 (1 k n) such that the matrix B = AP is
kk
lower-triangular.
(b) The principal minors of A are all different from 0.
Proof: Let P be any n n matrix and set B = AP. Then

B = A jr , P rk
jk
r
If P is upper-triangular and P = 1 for every k, then
kk
k 1
A P k > 1
jr rk = B – A ,
r 1 jk kk
Now B is lower-triangular provided B = 0 for j < k. Thus B will be lower-triangular if and only
jk
if
k 1
A P 1 j k – 1
jr rk = – A ,
r 1 kk
2 k n. ...(5)
So, we see that statement (a) in the lemma is equivalent to the statement that there exist scalars
P , 1 r k, 1 k n, which satisfy (5) and P = 1, 1 k n.
rk kk
In (5) for each k > 1 we have a system of k – 1 linear equations for the unknowns P , P , ..., P .
1k 2k k–1, k
The coefficient matrix of that system is
A n  A 1,k 1
 
A k 1  A k 1, k 1

and its determinant is the principal minor (A). If each (A) 0, the systems (5) have unique
k–1 k–1
solutions. We have shown that statement (b) implies statement (a) and that the matrix P is
unique.
Now suppose that (a) holds. Then, as we shall see,
(A) = (B)
k k
= B B ... B , k = 1, ..., n. ...(6)
11 22 kk
To verify (6), let A , ..., A and B , ... B be the columns of A and B, respectively. Then
1 n 1 n
B = A
1 1
r 1
B = P A j A r , r > 1. ...(7)
jr
r
j 1
Fix k, 1 k n. From (7) we see that the rth column of the matrix
B 11  B kk
 
B k 1  B kk

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