Page 203 - DMTH502_LINEAR_ALGEBRA

Unit 17: Invariant Direct Sums

If T is diagonalizable, T = c E + ... + c E , then Notes
1 1 k k
g(T) = g(c )E + ... + g(c )E
1 1 k k
for every polynomial g. Thus g(T) = 0 if and only if g(c ) = 0 for each i. In particular, the minimal
i
polynomial for T is
p = (x – c ) ... (x – c )
1 k
Now suppose T is a linear operator with minimal polynomial p = (x – c ) ... (x – c ), where c ,..., c
1 k 1 k
are distinct elements of the scalar field. We form the Lagrange polynomials
(x c )

j 
p = i
i j (c  c i )

j
So that p (c ) =  and for any polynomial g of degree less than or equal to (k – 1) we have
j i ij
g = g(c )p + ... + g(c )p
1 1 k k
Taking g to be the scalar polynomial 1 and then the polynomial x, we have
1  p   p 
1 k ...(2)
x  c p   c p k 
k 
1 1
You will note that the application to x may not be valid because k may be 1. But if k = 1, T is a
scalar multiple of the identity and hence diagonalizable). Now let E = p (T). From (2) we have
j j
I  E   E k 
1
 ...(3)
T  c E   c E
1 1 k k 
Observe that if i  j, then p p is divisible by the minimal polynomial p, because p p contains
i j i j
every (x – c ) as a factor. Thus
r
E E = 0, i  j ...(4)
i j
We must note one further thing, namely, that E  0 for each i. This is because p is the minimal
i
polynomial for T and so we cannot have p (T) = 0 since p has degree less than the degree of p. This
i i
last comment, together with (3), (4), and the fact that the c are distinct enables us to apply
i
Theorem 2 to conclude that T is diagonalizable.
Self Assessment
1. Let T be the diagonalizable linear operator on R which is represented by the matrix
3
 5  6  6
 
A   1 4 2
 
 3  6  4
 
use the Lagrange polynomials to write the representing matrix A in the form A = E + 2E ,
1 2
E + E = I, E E = 0. Where I is a unit matrix and 0 is zero matrix.
1 2 1 2
4
2. Let T be the linear operator on R which is represented by the 4 × 4 matrix
 0 1 0 1
 1 0 1 0 
A   
 0 1 0 1
 
 1 0 1 0 
Find the matrices E , E , E such that
1 2 3
A = C E + C E + C E , E + E + E = I and E E = 0 for i  j
1 1 2 2 3 3 1 2 3 i j

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