Page 202 - DMTH502_LINEAR_ALGEBRA
P. 202
Linear Algebra
Notes Certainly T is diagonalizable, since we have shown that every non-zero vector in the range of E
i
is a characteristic vector of T, and the fact that I = E + ... + E shows that these characteristic vectors
1 k
span V. All that remains to be demonstrated is that the null space of (T – c I) is exactly the range
i
of E . But this is clear, because if T = c , then
i i
k
(c c i )E 0
j
j
j 1
hence
(c – c )E = 0 for each j
j i j
and then
E = 0 j i
j
Since = E + ... + E , and E = 0 for j i, we have = E , which proves that is in the range
1 k j i
of E .
i
One part of Theorem 1 of unit 16 says that for a diagonalizable operator T, the scalars c ,..., c and
1 k
the operators E ,..., E are uniquely determined by conditions (i), (ii), (iii), the fact that the c are
1 k i
distinct, and the fact that the E are non-zero. One of the pleasant features of the decomposition
i
T = c E + ... + c E is that if g is any polynomial over the field F, then
1 1 k k
g(T) = g(c )E + ... + g(c )E .
1 1 k k
To see how it is proved one need only compute T for each positive integer r. For example,
r
k k
T = c E i c E
2
i j j
i 1 j 1
k k
= c c E E j
i
i j
i 1 j 1
k
2
= c E 2 i
i
i 1
k
2
= c E i
i
i 1
The reader should compare this with g(A) where A is a diagonal matrix; for then g(A) is simply
the diagonal matrix with diagonal entries g(A ), ..., g(A ).
11 nn
We should like in particular to note what happens when one applies the Lagrange polynomials
corresponding to the scalars c ,..., c :
1 k
(x c )
j
p = i
i j (c c i )
j
We have p (c ) = , which means that
j i ij
k
p (T) = E
j ij i
i 1
= E
j
Thus the projections E not only commute with T but are polynomials in T.
j
Such calculations with polynomials in T can be used to give an alternative proof of Theorem 2 of
unit 14, which characterized diagonalizable operators in terms of their minimal polynomials.
The proof is entirely independent of our earlier proof.
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