Page 178 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes since f is the determinant of the matrix xI – A whose entries are the polynomials
(xI A ) x A .
ij ij ji
We wish to show that f(T) = 0. In order that f(T) be the zero operator, it is necessary and sufficient
that (det B) = 0 for k = 1, , n. By the definition of B, the vectors  ,   satisfy the equations
k 1 n
n
B ij j 0, 1 i . n … (3)
j 1
When n = 2, it is suggestive to write (3) in the form

T A I A I 1 0 .
21
11
A I T A I 0
12 22 2
In this case, the classical adjoint, adj B is the matrix
22
21
B  T A I A I
A I T A I
12 11
and

BB detB 0 .
0 detB
Hence, we have


(det ) 1 = (BB ) 1
B
2 2

= B B 1
2
0
= 0 .


In the general case, let B = adj B. Then by (3)
n

B B ij j 0
ki
j 1
for each pair k, i, and summing on i, we have

n n

B B
0 = ki ij i
i 1 j 1
n n 
= B B ij j .
ki
j 1 i 1

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