Suppose S is in L(V). Define
{.,.}_1 by {u, v}_1 = {Su, Sv} for all u, v in V such that {.,.} is an inner product on V if and only if S is injective

To address this problem, let's break it down into key components and provide a clear explanation.

### Problem Breakdown

You have:
- A vector space \( V \) over some field \( \mathbb{F} \).
- An inner product \( \{ \cdot, \cdot \} \) defined on \( V \).
- A linear map \( S: V \to V \), which is an element of \( L(V) \), the space of linear operators on \( V \).

You are defining a new function \( \{ \cdot, \cdot \}_1 \) on \( V \) by:
\[
\{ u, v \}_1 = \{ Su, Sv \}
\]
for all \( u, v \in V \).

### Question

We need to determine under what conditions \( \{ \cdot, \cdot \}_1 \) is an inner product on \( V \). Specifically, the statement suggests that \( \{ \cdot, \cdot \}_1 \) is an inner product on \( V \) if and only if \( S \) is injective.

### Solution

#### Inner Product Definition

For \( \{ \cdot, \cdot \}_1 \) to be an inner product on \( V \), it must satisfy the following conditions:

1. **Bilinearity**: 
   - \( \{ au + bw, v \}_1 = a\{ u, v \}_1 + b\{ w, v \}_1 \)
   - \( \{ u, av + bw \}_1 = a\{ u, v \}_1 + b\{ u, w \}_1 \)
   - This is true because the original inner product \( \{ \cdot, \cdot \} \) is bilinear, and \( S \) is a linear map. Therefore, bilinearity is preserved.

2. **Symmetry**: 
   - \( \{ u, v \}_1 = \{ v, u \}_1 \)
   - Since the original inner product is symmetric (\( \{ u, v \} = \{ v, u \} \)) and \( S \) is a linear operator, symmetry is preserved.

3. **Positive Definiteness**:
   - \( \{ u, u \}_1 > 0 \) for all non-zero \( u \in V \), and \( \{ u, u \}_1 = 0 \) if and only if \( u = 0 \).

Let's focus on this third condition, as it is the most critical for the problem.

#### Positive Definiteness Condition

Given \( \{ u, u \}_1 = \{ Su, Su \} \), for \( \{ \cdot, \cdot \}_1 \) to be positive definite, we require that:
\[
\{ u, u \}_1 = 0 \implies u = 0
\]

However, \( \{ u, u \}_1 = \{ Su, Su \} = 0 \) if and only if \( Su = 0 \). For this to imply that \( u = 0 \), \( S \) must be **injective**. In other words, the kernel of \( S \) must be trivial (i.e., \( \text{ker}(S) = \{0\} \)).

If \( S \) is not injective, there exists some non-zero \( u \) such that \( Su = 0 \), which would lead to \( \{ u, u \}_1 = \{ Su, Su \} = 0 \) even though \( u \neq 0 \). This violates the positive definiteness condition of the inner product.

#### Conclusion

Thus, \( \{ \cdot, \cdot \}_1 \) is an inner product on \( V \) if and only if \( S \) is injective.

### Summary

The function \( \{ \cdot, \cdot \}_1 \) defined by \( \{ u, v \}_1 = \{ Su, Sv \} \) is an inner product on \( V \) if and only if the linear operator \( S \) is injective.

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Would you like to go into more detail about any part of this explanation or have any questions?

Here are 5 related questions to consider:
1. What are the necessary and sufficient conditions for a linear map to be injective?
2. How does the concept of orthogonality change under the new inner product \( \{ \cdot, \cdot \}_1 \)?
3. Can a linear operator that is not injective ever induce a non-degenerate bilinear form?
4. How does the determinant of \( S \) relate to the injectivity of \( S \)?
5. What is the geometric interpretation of \( S \) when \( \{ \cdot, \cdot \}_1 \) is an inner product?

**Tip:** Always check the properties of linear operators when dealing with induced inner products. The injectivity of an operator is often key in preserving properties like positive definiteness.

Suppose S is in L(V). Define {.,.}_1 by {u, v}_1 = {Su, Sv} for all u, v in V such that {.,.} is an inner product on V if and only if S is injective.

Explore the conditions under which the function {u, v}_1 = {Su, Sv} forms an inner product on a vector space V. This problem delves into key concepts in linear algebra, including inner products, linear maps, and the importance of injectivity for positive definiteness. Ideal for undergraduate students studying advanced linear algebra.

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