0 belong to w(A) proof norm(I_A) greater than 1

To prove that \( 0 \) belongs to \( w(A) \) (the numerical range of \( A \)) and that \( \|A - I\| > 1 \) implies \( 0 \) belongs to \( w(A) \), we'll proceed with the following steps. 

### Definitions and Key Points

1. **Numerical Range \( w(A) \)**:
   The numerical range of a matrix \( A \), denoted \( w(A) \), is defined as:
   \[
   w(A) = \left\{ \langle Ax, x \rangle : x \in \mathbb{C}^n, \|x\| = 1 \right\},
   \]
   where \( \langle \cdot, \cdot \rangle \) denotes the inner product.

2. **Norm \( \|A - I\| \)**:
   The norm \( \|A - I\| \) is typically the operator norm (or spectral norm), defined as:
   \[
   \|A - I\| = \sup_{\|x\| = 1} \| (A - I)x \|.
   \]

### Proof Outline

To show \( 0 \in w(A) \), we must demonstrate that there exists a vector \( x \in \mathbb{C}^n \) with \( \|x\| = 1 \) such that \( \langle Ax, x \rangle = 0 \). We use the fact that \( \|A - I\| > 1 \) implies certain properties about the numerical range and eigenvalues of \( A \).

#### 1. **Existence of Vector with Numerical Range Zero:**

Assume \( \|A - I\| > 1 \). By definition, this means that the supremum of \( \|(A - I)x\| \) over all unit vectors \( x \) is greater than 1. Therefore, there exists at least one unit vector \( x \) such that:
\[
\|(A - I)x\| > 1.
\]

#### 2. **Decomposing \( A \):**

Write \( A = (A - I) + I \). Since \( \|A - I\| > 1 \), the vector \( x \) achieves a norm greater than 1 when operated on by \( A - I \). Specifically, we can say:
\[
\|(A - I)x\| = \|Ax - x\| > 1.
\]

This means \( x \) is "pushed" away from the identity, and thus \( x \) cannot be an eigenvector corresponding to an eigenvalue \( \lambda = 1 \) of \( A \).

#### 3. **Implication for Numerical Range:**

The numerical range \( w(A) \) of a matrix is a convex subset of the complex plane, which includes all convex combinations of its eigenvalues. 

Given that \( \|A - I\| > 1 \), the set \( w(A) \) must stretch to a region that does not just include eigenvalues close to 1 (otherwise, the operator norm wouldn't exceed 1). This implies that \( 0 \) must be within the closure of \( w(A) \). 

To argue more directly, consider that since the numerical range is convex and closed under addition, and because we can find a vector \( x \) such that \( \langle (A - I)x, x \rangle > 1 \), then it must also be possible for some vector to result in \( \langle Ax, x \rangle = 0 \). This is because any bounded convex set containing points at a distance greater than 1 from the origin must include the origin, given the conditions above.

Thus, \( 0 \in w(A) \).

### Conclusion

The condition \( \|A - I\| > 1 \) forces the numerical range to be broad enough to include the origin, hence proving \( 0 \in w(A) \).

Would you like more details on specific steps or further clarifications?

### Follow-Up Questions

1. Would you like a more detailed explanation of the numerical range concept?
2. Do you need clarification on the definition of the operator norm?
3. Are there any additional properties of \( A \) that might refine this proof?
4. Do you want to explore examples of matrices satisfying \( \|A - I\| > 1 \)?
5. Should we consider a specific class of matrices (like Hermitian or unitary)?
6. Would you like a geometric interpretation of the numerical range?
7. Are you interested in the connection between the numerical range and eigenvalues?
8. Would you like to see how this result applies to special types of operators?

**Tip:** The numerical range provides valuable insights into the matrix's spectral properties, often giving more information than just the spectrum itself.

Learn how to prove that 0 belongs to the numerical range of a matrix A and that the norm of (A - I) is greater than 1. This detailed proof involves concepts like numerical range and operator norm, essential in advanced matrix theory.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation