1/3
5.A.34
By definition, 8x + null T 2 V /null T . We have
T /null T (x + null T ) = Tx + null T
Then T /null T is injective is equivalent to
x 2 null T , null T = x + null T , Tx + null T = null T , Tx 2 null T
Remark 1. it can be viewed as l.h.s is x = 0, r.h.s is Tx = 0.
Then Tx 2 null T , x 2 null T .
Suppose x 2 null T \ range T , then 9v; Tv = x ) Tv 2 null T ) Tv = 0.
Similarly, if null T \ range T = f0g, suppose x 2 null T ) x = 0 ) Tx = 0 2 null T .
Suppose Tx 2 null T also notice that Tx 2 range T , thus x = 0 ) x 2 null T .
2/3
5.A.36
Take 1; x; x2; : : : l.i in PR. Consider V = P (R); U = P (R) ¡ P1(R).
Define Tp = xp. We can see that T is linear and U is invariant under T .
Then
(T /U )(1 + U ) = x + U = 0 + U
Since 1 + U =
/ 0 + U , thus 0 is an eigenvalue of T /U with eigenvector 1 + U .
But 0 is not a eigenvalue of T , assume 9p 0 =
/ 0 2 P (R) s.t xp = 0, which is impossible since
x
/ 0 on R.
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5.B.3
Idea: Split V into two eigenspace null(T + I) and null(T ¡ I).
Thus
1
1
1
1
v = (v ¡ Tv) + (v + Tv) ) (T + I) (v ¡ Tv) = (T + I) (v + Tv) = 0
2
2
2
2
V = null(T + I) + null(T ¡ I)
Since ¡1 is not an eigenvalue of T , then null(T + I) = 0, otherwise T + I is not surjective thus
¡1 is an eigenvector (5.6).
Hence V = null(T ¡ I) ) 8v 2 V ; (T ¡ I)v = 0 ) T = I