Page 168 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes for its diagonal entries the scalars c , each repeated a certain number of times. If c is repeated d
i i i
times, then (we may arrange that) the matrix has the block form
c I 0 ... 0
1 1
0 c I ... 0
2 2
T
[ ]    
0 0 ... c I
k k
where I is the d × d identity matrix. From that matrix we see two things. First, the characteristic
j j j
polynomial for T is the product of (possibly repeated) linear factors:
d
f = (x – c ) 1 ... (x – c ) k
d
1 k
If the scalar field F is algebraically closed, e.g., the field of complex numbers, every polynomial
over F can be so factored; however, if F is not algebraically closed, we are citing a special
property of T when we say that its characteristic polynomial has such a factorization. The second
thing we see that d , the number of times which c is repeated as root of f, is equal to the
i i
dimension of the space of characteristic vectors associated with the characteristic value c . That is
i
because the nullity of a diagonal matrix is equal to the number of zeros which it has on its main
diagonal, and the matrix [T – c I] has d zeros on its main diagonal. This relation between the
i  i
dimension of the characteristic space and the multiplicity of the characteristic value as a root of
f does not seem exciting at first; however, it will provide us with a simpler way of determining
whether a given operator is diagonalizable.
Lemma: Let T be a linear operator on the finite dimensional space V. Let c , ..., c be the distinct
1 k
characteristic values of T and let W be the space of characteristic vectors associated with the
i
characteristic value c . If W = W + ... + W , then
i i k
dim W = dim W + ... + dim W .
1 k
In fact if B is an ordered basis for W , then  = ( , ...,  ) is an ordered basis for W.
i i 1 k
Proof: The space W = W + ... + W is the subspace spanned by all of the characteristic vectors of T.
i k
Usually when one forms the sum W of subspaces W , one expects that dim W < dim W + ... + dim
i i
W because of linear relations which may exist between vectors in the various spaces. This
k
lemma states that the characteristic spaces associated with different characteristic values are
independent of one another.
Suppose that (for each i) we have a vector in W , and assume that + ... + = 0. We shall show
i i i k
that = 0 for each i. Let f be any polynomial. Since T = c , the preceding lemma tells us that
i i i i
0 = f(T)0 = f(T) + ... + f(T)
1 k
= f(c ) + ... + f(c )
1 1 k k
Choose polynomial f , ..., f such that
1 k
1, i j
f (c ) = ij
i j 0, i . j
Then

0 = f (T) = ij j
i
j
= .
i
Now, let be an ordered basis for W , and let be the sequence = ( , ..., ). Then spans the
i i 1 k
subspace W = W + ... + W . Also, is a linearly independent sequence of vectors, for the following
1 k
reason. Any linear relation between the vectors in will have the form + ... + = 0, where
1 k i

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