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Sagot :
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.
### (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.
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