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Let [tex]T[/tex] be a linear operator on an inner product space [tex]V[/tex], and suppose that [tex]\|T(x)\| = \|x\|[/tex] for all [tex]x \in V[/tex]. Prove that [tex]T[/tex] is one-to-one.

Sagot :

GinnyAnswer
To prove that the linear operator [tex]\( T \)[/tex] on an inner product space [tex]\( V \)[/tex] is one to one, given that [tex]\( \|T(x)\| = \|x\| \)[/tex] for all [tex]\( x \in V \)[/tex], we need to demonstrate that if [tex]\( T(x) = 0 \)[/tex], then [tex]\( x = 0 \)[/tex]. This property would ensure that [tex]\( T \)[/tex] is injective (one to one).

### Step-by-Step Solution

1. Given:
- [tex]\( T \)[/tex] is a linear operator on an inner product space [tex]\( V \)[/tex].
- [tex]\( \|T(x)\| = \|x\| \)[/tex] for all [tex]\( x \in V \)[/tex].

2. Objective:
- Prove that [tex]\( T \)[/tex] is one to one. That is, prove if [tex]\( T(x) = 0 \)[/tex], then [tex]\( x = 0 \)[/tex].

3. Proof:
- Assume [tex]\( T(x) = 0 \)[/tex] for some [tex]\( x \in V \)[/tex].

4. Apply the given condition:
- Since [tex]\( T \)[/tex] is an isometry, we have [tex]\( \|T(x)\| = \|x\| \)[/tex] for all [tex]\( x \in V \)[/tex].

5. Substitute [tex]\( T(x) = 0 \)[/tex] into the given condition:
- If [tex]\( T(x) = 0 \)[/tex], then [tex]\( \|T(x)\| = \|0\| \)[/tex].

6. Evaluate the norm of [tex]\( 0 \)[/tex]:
- [tex]\( \|0\| = 0 \)[/tex].

7. Thus:
- [tex]\( \|T(x)\| = 0 \)[/tex].

8. By the given condition:
- [tex]\( \|T(x)\| = \|x\| \)[/tex].

9. So:
- [tex]\( \|x\| = 0 \)[/tex].

10. Interpret the norm of [tex]\( x \)[/tex]:
- In an inner product space, the norm of a vector is zero if and only if the vector itself is zero.

11. Conclusion:
- Since [tex]\( \|x\| = 0 \)[/tex], it implies [tex]\( x = 0 \)[/tex].

12. Final result:
- We have shown that if [tex]\( T(x) = 0 \)[/tex], then [tex]\( x = 0 \)[/tex].
- Therefore, [tex]\( T \)[/tex] is one to one.

In conclusion, we have rigorously proven that the linear operator [tex]\( T \)[/tex] is one to one based on the initial condition that [tex]\( \|T(x)\| = \|x\| \)[/tex] for all [tex]\( x \in V \)[/tex].
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