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.