Prove the following for a normal operator [tex]\( T \)[/tex]:

1. [tex]\( T(v) = 0 \)[/tex] if and only if [tex]\( T^*(v) = 0 \)[/tex].
2. [tex]\( T - \lambda I \)[/tex] is normal, where [tex]\( \lambda \)[/tex] is a scalar.



Answer :

Alright! Let's prove the given statements step-by-step.

### (7.1) [tex]\( T(v) = 0 \)[/tex] if and only if [tex]\( T^(v) = 0 \)[/tex]

To prove this, we need to show two implications:

1. If [tex]\(T(v) = 0\)[/tex], then [tex]\(T^
(v) = 0\)[/tex].
2. If [tex]\(T^(v) = 0\)[/tex], then [tex]\(T(v) = 0\)[/tex].

#### Implication 1: [tex]\( T(v) = 0 \implies T^
(v) = 0 \)[/tex]

Assume [tex]\( T(v) = 0 \)[/tex]. We want to show that [tex]\( T^(v) = 0 \)[/tex].

Let [tex]\( u \)[/tex] be any vector in the Hilbert space.

[tex]\[ \langle T(v), u \rangle = 0 \][/tex]

Since [tex]\( T(v) = 0 \)[/tex],

[tex]\[ \langle 0, u \rangle = 0 \][/tex]

Using the property of the inner product, [tex]\( \langle 0, u \rangle = 0 \)[/tex] for all [tex]\( u \)[/tex]. Therefore, we have:

[tex]\[ \langle v, T^
(u) \rangle = 0 \][/tex]

Because [tex]\( u \)[/tex] was arbitrary, this implies [tex]\( v \)[/tex] is orthogonal to every vector of the form [tex]\( T^(u) \)[/tex]. The only vector orthogonal to a dense subset in a Hilbert space is the zero vector. Thus,

[tex]\[ T^
(v) = 0 \][/tex]

#### Implication 2: [tex]\( T^(v) = 0 \implies T(v) = 0 \)[/tex]

Assume [tex]\( T^
(v) = 0 \)[/tex]. We want to show that [tex]\( T(v) = 0 \)[/tex].

Let [tex]\( u \)[/tex] be any vector in the Hilbert space.

[tex]\[ \langle v, T^(u) \rangle = 0 \][/tex]

Since [tex]\( T^
(v) = 0 \)[/tex],

[tex]\[ \langle v, 0 \rangle = 0 \][/tex]

Using the property of the inner product, [tex]\( \langle v, 0 \rangle = 0 \)[/tex] for all [tex]\( u \)[/tex]. Therefore, we have:

[tex]\[ \langle T(v), u \rangle = 0 \][/tex]

Because [tex]\( u \)[/tex] was arbitrary, this implies [tex]\( T(v) \)[/tex] is orthogonal to every vector. The only vector orthogonal to a dense subset in a Hilbert space is the zero vector. Thus,

[tex]\[ T(v) = 0 \][/tex]

This completes the proof for [tex]\( T(v) = 0 \)[/tex] if and only if [tex]\( T^(v) = 0 \)[/tex].

### (7.2) [tex]\( T - \lambda I \)[/tex] is normal, where [tex]\( \lambda \)[/tex] is a scalar

To prove this, we need to show that [tex]\( (T - \lambda I) \)[/tex] is normal if [tex]\( T \)[/tex] is normal. Recall that an operator [tex]\( T \)[/tex] is normal if [tex]\( T T^
= T^ T \)[/tex].

First, note that [tex]\( T - \lambda I \)[/tex] is the operator [tex]\( T \)[/tex] shifted by [tex]\( \lambda \)[/tex], where [tex]\( I \)[/tex] is the identity operator.

Given that [tex]\( T - \lambda I \)[/tex]'s adjoint is [tex]\( T^
- \bar{\lambda} I \)[/tex]:

[tex]\[ (T - \lambda I)^ = T^ - \bar{\lambda}I \][/tex]

Let's show that [tex]\( (T - \lambda I) (T^ - \bar{\lambda} I) = (T^ - \bar{\lambda} I) (T - \lambda I) \)[/tex].

Expanding both sides, we get:

Left Side (LS):
[tex]\[ (T - \lambda I) (T^ - \bar{\lambda} I) = T T^ - T \bar{\lambda} I - \lambda I T^ + \lambda \bar{\lambda} I \][/tex]

Right Side (RS):
[tex]\[ (T^
- \bar{\lambda} I) (T - \lambda I) = T^ T - T^ \lambda I - \bar{\lambda} I T + \bar{\lambda} \lambda I \][/tex]

Notice:
[tex]\[ \lambda \bar{\lambda} = \bar{\lambda} \lambda \][/tex]
[tex]\[ T \bar{\lambda} I = \bar{\lambda} T I = \bar{\lambda} T \][/tex]
[tex]\[ \lambda I T^ = \lambda T^ I = \lambda T^ \][/tex]

Given that [tex]\(T\)[/tex] is normal ([tex]\(T T^
= T^ T\)[/tex]):

[tex]\[ LS = T T^
- \lambda T^ - \bar{\lambda} T + \lambda \bar{\lambda} I \][/tex]
[tex]\[ RS = T^
T - \lambda T^ - \bar{\lambda} T + \lambda \bar{\lambda} I \][/tex]

Since [tex]\( T T^
= T^ T \)[/tex],

[tex]\[ LS = TS = RS \][/tex]

Hence, [tex]\( LS = RS \)[/tex], proving that [tex]\( (T - \lambda I) (T^
- \bar{\lambda} I) = (T^* - \bar{\lambda} I) (T - \lambda I) \)[/tex].

Therefore, [tex]\( T - \lambda I \)[/tex] is normal.

This completes the proof for [tex]\( T - \lambda I \)[/tex] being normal.