Page 201 - DMTH502_LINEAR_ALGEBRA

Unit 17: Invariant Direct Sums

= TE  Notes
j
This holds for each  in V, so E T = TE .
j j
We shall now describe a diagonalizable operator T in the language of invariant direct sum
decompositions (projections which commute with T). This will be a great help to us in
understanding some deeper decomposition theorems later. The description which we are about
to give is rather complicated, in comparison to the matrix formulation or to the simple statement
that the characteristic vectors of T span the underlying space. But, we should bear in mind that
this is our first glimpse at a very effective method, by means of which various problems concerned
with subspaces, bases, matrices, and the like can be reduced to algebraic calculations with linear
operators. With a little experience, the efficiency and elegance of this method of reasoning
should become apparent.
Theorem 2: Let T be a linear operator on a finite-dimensional space V. If T is diagonalizable and
if c ,..., c are the distinct characteristic values of T, then there exist linear operators E ,..., E on V
1 k 1 k
such that
(i) T = c E + ... + c E ;
1 1 k k
(ii) I = ‘E + ... + E ;
1 k
(iii) E E = 0, i  j;
i j
2
(iv) E  E (E is a projection);
1 i i
(v) the range of E is the characteristic space for T associated with c .
i i
Conversely, if there exist k distinct scalars c ,..., c and k non-zero linear operators E ,..., E which
1 k 1 k
satisfy conditions (i), (ii), and (iii), then T is diagonalizable, c ,..., c are the distinct characteristic
1 k
values of T, and conditions (iv) and (v) are satisfied also.
Proof: Suppose that T is diagonalizable, with distinct characteristic values c ,..., c . Let W be the
1 k i
space of characteristic vectors associated with the characteristic value c . As we have seen,
i
V = W  ...  W
1 k
Let E ,...,E be the projections associated with this decomposition, as in Theorem 1 of unit 16.
1 k
Then (ii), (iii), (iv) and (v) are satisfied. To verify (i), proceed as follows. For each  in V,
 = E  + ... + E 
1 k
and so
T = TE  + ... + TE 
1 k
= c E  + ... + c E 
1 1 k k
In other words, T = c E + ... + c E .
1 1 k k
Now suppose that we are given a linear operator T along with distinct scalars c and non-zero
i
operators E which satisfy (i), (ii) and (iii). Since E E = 0 when i  j, we multiply both sides of
i i j
2
I = E + ... + E by E and obtain immediately E  E . Multiplying T = c E + ... + c E by E , we then
1 k i i i 1 1 k k i
have TE = c E , which shows that any vector in the range of E is in the null space of (T – c I). Since
i i i i i
we have assumed that E  0, this proves that there is a non-zero vector in the null space of
i
(T – c I), i.e., that c is a characteristic value of T. Furthermore, the c are all of the characteristic
i i i
values of T; for, if c is any scalar, then
T – cI = (c – c)E + ... + (c – c)E
1 1 k k
so if (T – cI) = 0, we must have (c – c)E  = 0. If  is not the zero vector, then E   0 for some i,
i i i
so that for this i we have c – c = 0.
i
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