Page 166 - Linear Algebra Done Right

Chapter 7. Operators on Inner-Product Spaces
154
√
We see from 7.43 that S 1 maps range T T onto range T. Clearly
In the rest of the proof
∗
√
all we are doing is
∗
extending S 1 to an 7.42 and 7.43 imply that S 1 u = u for all u ∈ range T T.In
particular, S 1 is injective. Thus from 3.4, applied to S 1 , we have
isometry S on all of V. √
dim range T T = dim range T.
∗
√
This implies that dim(range T T) = dim(range T) (see Exercise 15
⊥
⊥
∗
√
in Chapter 6). Thus orthonormal bases (e 1 ,...,e m ) of (range T T) ⊥
∗
and (f 1 ,...,f m ) of (range T) can be chosen; the key point here is that
⊥
these two orthonormal bases have the same length. Define a linear map
√
S 2 : (range T T) → (range T) by
∗
⊥
⊥
S 2 (a 1 e 1 +· · ·+ a m e m ) = a 1 f 1 +· · ·+ a m f m .
√
Obviously S 2 w = w for all w ∈ (range T T) .
⊥
∗
√
Now let S be the operator on V that equals S 1 on range T T and
∗
√
equals S 2 on (range T T) . More precisely, recall that each v ∈ V
⊥
∗
can be written uniquely in the form
7.44 v = u + w,
√ √
where u ∈ range T T and w ∈ (range T T) (see 6.29). For v ∈ V
⊥
∗
∗
with decomposition as above, define Sv by
Sv = S 1 u + S 2 w.
For each v ∈ V we have
√ √
S( T Tv) = S 1 ( T Tv) = Tv,
∗
∗
√
so T = S T T, as desired. All that remains is to show that S is an isom-
∗
etry. However, this follows easily from the two uses of the Pythagorean
theorem: if v ∈ V has decomposition as in 7.44, then
2 2
Sv = S 1 u + S 2 w
2 2
= S 1 u + S 2 w
2 2
= u + w
2
= v ,
where the second equality above holds because S 1 u ∈ range T and
S 2 u ∈ (range T) .
⊥