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].
