Chapter 10: Problem 19
Sei \(B=\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) eine Basis des euklidischen Vektorraums \(V\). Zeigen Sie: Eine lineare Abbildung von \(V\) in sich ist genau dann selbstadjungiert, wenn \(\left\langle f\left(v_{i}\right), v_{j}\right\rangle=\left\langle v_{i}, f\left(v_{j}\right)\right\rangle\) für alle \(i, j \in\\{1,2, \ldots, n\\}\) gilt.
Short Answer
Step by step solution
Understand the Problem
Analyze the Definition of Self-adjoint
Express with Basis Vectors
Check the Given Condition
Prove the 'If' Part
Prove the 'Only If' Part
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Self-Adjoint
- \( \left\langle f(x), y \right\rangle = \left\langle x, f(y) \right\rangle \)
Inner Product Space
- It is linear in its first argument: \( \left\langle au + bv, w \right\rangle = a\left\langle u, w \right\rangle + b\left\langle v, w \right\rangle \)
- It is conjugate symmetric: \( \left\langle u, v \right\rangle = \overline{\left\langle v, u \right\rangle} \)
- The inner product of a vector with itself is always non-negative: \( \left\langle u, u \right\rangle \geq 0 \)
- It is zero if and only if the vector itself is the zero vector: \( \left\langle u, u \right\rangle = 0 \) implies \( u = 0 \)
Eigenvectors
- \( Av = \lambda v \)
Symmetric Matrix
- \( A = A^T \)