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Linear Algebra

Notes across with two problems. First, T may not have a single characteristic value due to the limitation
of the scalar field. Second, even if the characteristic polynomial factors completely over F into a
product of polynomials of degree 1, there may not be enough characteristic vectors for T to span
the space V; which is clearly a deficiency in T. The second situation is illustrated by the operator
3
T on F (F any field) represented in the standard basis by
 2 0 0 
 
A  1 2 0
 
 0 0  1
 
The characteristic polynomial for A is (x – 2) (x + 1) and this is plainly also the minimal polynomial
2
for A (or for T). Thus T is not diagonalizable. One sees that this happens because the null space
of (T – 2I) has dimension 1 only. On the other hand, the null space of (T + I) and the null space of
(T – 2I) together span V, the former being the subspace spanned by  and the latter the subspace
2
3
spanned by  and  .
1 2
This will be more or less our general method for the second problem. If (remember this is an
assumption) the minimal polynomial for T decomposes
r
r
p = (x – c ) 1 ... (x – c ) 2
1 k
where c ,..., c are distinct elements of F, then we shall show that the space V is the direct sum of
1 k
the null spaces of (T – c I) , i = 1,..., k. The hypothesis about p is equivalent to the fact that T is
ri
i
triangulable (Theorem 1 of unit 14); however, that knowledge will not help us.
The theorem which we prove is more general than what we have described, since it works with
the primary decomposition of the minimal polynomial, whether or not the primes which enter
are all of first degree. The reader will find it helpful to think of the special case when the primes
are of degree 1, and even more particularly, to think of the projection-type proof of Theorem 2
of unit 14, a special case of this theorem.

18.2 Primary Decomposition Theorem

Theorem 1 (Primary Decomposition Theorem): Let T be a linear operator on the finite-dimensional
vector space V over the field F. Let p be the minimal polynomial for T,

1 r
p = p  p r k
1
where the p are distinct irreducible monic polynomials over F and the r are positive integers.
i i
Let W be the null space of p (T) i , i = 1,..., k. Then
r
i i
...
(i) V = W   W ;
1 k
(ii) each W is invariant under T;
i
(iii) if T is the operator induced on W by T, then the minimal polynomial for T is p 1 r .
i i i 1
Proof: The idea of the proof is this. If the direct-sum decomposition (i) is valid, how can we get
hold of the projections E ,..., E associated with the decomposition? The projection E will be the
1 k i
identity on W and zero on the other W . We shall find a polynomial h such that h (T) is the
i j i i
...
identity on W and is zero on the other W , and so that h (T) + + h (T) = I, etc.
i j 1 k
For each i, let
p
f  i r  p i r .

i j
p

i j i
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