Page 208 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes and in general that
r
r
N = (T – c I) E + ... + (T – c I) E .
r
1 1 k k
When r  r for each i, we shall have N = 0, because the operator (T – c I) will then be 0 on the
r
r
i i
range of E .
i
Definition: Let N be a linear operator on the vector space V. We say that N is nilpotent if there is
some positive integer r such that N = 0.
r
Theorem 2: Let T be a linear operator on the finite-dimensional vector space V over the field F.
Suppose that the minimal polynomial for T decomposes over F into a product of linear polynomials.
Then there is a diagonalizable operator D on V and a nilpotent operator N on V such that
(i) T = D + N,
(ii) DN = ND
The diagonalizable operator D and the nilpotent operator N are uniquely determined by (i) and
(ii) and each of them is a polynomial in T.
Proof: We have just observed that we can write T = D + N where D is diagonalizable and N is
nilpotent, and where D and N not only commute but are polynomials in T. Now suppose that we
also have T = D’ + N’ where D’ is diagonalizable, N’ is nilpotent, and D’N’ = N’D’. We shall prove
that D = D’ and N = N’.
Since D’ and N’ commute with one another and T = D’ + N’, we see that D’ and N’ commute with
T. Thus D’ and N’ commute with any polynomial in T; hence they commute with D and with N.
Now we have
D + N = D’ + N’
or

D – D’ = N’ – N
and all four of these operators commute with one another. Since D and D’ are both diagonalizable
and they commute, they are simultaneously diagonalizable, and D – D’ is diagonalizable. Since
N and N’ are both nilpotent and they commute, the operator (N’ – N) is nilpotent; for, using the
fact that N and N’ commute

r r  

r
N
( ' N )     ( ') r j ( N ) j


N
j 0  j
and so when r is sufficiently large every term in this expression for (N’ – N) will be 0. (Actually,
r
a nilpotent operator on an n-dimensional space must have its nth power 0; if we take r = 2n
above, that will be large enough. It then follows that r = n is large enough, but this is not obvious
from the above expression.) Now D – D’ is a diagonalizable operator which is also nilpotent.
Such an operator is obviously the zero operator; for since it is nilpotent, the minimal polynomial
for this operator is of the form x for some r  m; but then since the operator is diagonalizable, the
r
minimal polynomial cannot have a repeated root; hence r = 1 and the minimal polynomial is
simply x, which says the operator is 0. Thus we see that D = D’ and N = N’.
Corollary: Let V be a finite-dimensional vector space over an algebraically closed field F, e.g.,
the field of complex numbers. Then every linear operator T on V can be written as the sum of a
diagonalizable operator D and a nilpotent operator N which commute. These operators D and N’
are unique and each is a polynomial in T.
From these results, one sees that the study of linear operators on vector spaces over an
algebraically closed field is essentially reduced to the study of nilpotent operators. For vector

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