Page 176 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes = (T – cI)
and thus, c is a characteristic value of T.
Now, suppose that c is a characteristic value of T, say T = c with 0. So

T
p ( ) = ( ) .p c
Since p(T) = 0 and 0, we have p(c) 0.

Let T be a diagonalizable linear operator and let c 1 , , c be the distinct characteristic values of
k
T. Then it is easy to see that the minimal polynomial for T is the polynomial.
p = (x – c )  (x – c ).
1 k
If  is a characteristic vector, then one of the operators T – c I,  , T – c I sends  into 0. Therefore
1 k

(T c I ) (T c I ) 0
1 k
for every characteristic vector . There is a basis for the underlying space which consists of
characteristic vectors of T; hence

T
p ( ) (T c I ) (T c I ) 0.
1 k
What we have concluded is this. If T is a diagonalizable linear operator, then the minimal
polynomial for T is a product of distinct linear factors. As we shall soon see, that property
characterizes diagonalizable operators.

Example 1: Let’s try to find the minimal polynomials for the operators in Example 1, 2,
and 6 in unit 12. We shall discuss them in reverse order. The operator in Example 6 was found to
be diagonalizable with characteristic polynomial.
From the preceding paragraph we know that the minimal polynomial for T is.
p (x 1)(x x ).
The reader might find it reassuring to verify directly that
I
(A I )(A 2 ) 0.
2
In Example 2, the operator T also had the characteristic polynomial f (x 1)(x 2) . But, this T
is not diagonalizable, so we don’t know that the minimal polynomial is (x – 1) (x – 2). What do
be know about the minimal polynomial in this case? We know that its roots are 1 and 2, with
some multiplicities allowed. Thus we search for p among polynomials of the form (x – 1) k
l
(x 2) ,k 1,l 1. Try (x – 1) (x – 2):
2 1 1 1 1 1
2 1 1 2 0 1
(A – I) (A – 2I) =
2 2 1 2 2 2
2 0 1
= 2 0 1
4 0 2
Thus, the minimal polynomial has degree at least 3. So, next we should try either (x – 1) (x – 2)
2
or (x – 1) (x – 2) . The second being the characteristic polynomial, would seem a less random
2
choice. One can readily compute that (A – I) (A – 2I) = 0. Thus the minimal polynomial for T is
2
its characteristic polynomial.

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