All eigenvalues of a symmetric operator are real
Dependencies:
- Symmetric operator
- Eigenvalues and Eigenvectors
- Inner product is anti-linear in second argument
- A field is an integral domain
Let $L: V \mapsto V$ be a symmetric operator, where $V$ is a vector space over a subfield of $\mathbb{C}$.
All eigenvalues of $L$ are real.
Proof
Let $(\lambda, u)$ be an eigenvalue-eigenvector pair of $L$.
\begin{align} 0 &= \langle L(u), u \rangle - \langle u, L(u) \rangle \tag{$L$ is symmetric} \\ &= \langle \lambda u, u \rangle - \langle u, \lambda u \rangle \\ &= \lambda \langle u, u \rangle - \overline{\lambda} \langle u, u \rangle \tag{(anti-)linearity} \\ &= (\lambda - \overline{\lambda}) \langle u, u \rangle \end{align}
Since $u$ is an eigenvector, $u \neq 0$. By definiteness of inner-product, $\langle u, u \rangle \neq 0$. Since $\mathbb{C}$ is a field, it has no zero-divisors. Therefore, \[ (\lambda - \overline{\lambda}) \langle u, u \rangle = 0 \implies \lambda = \overline{\lambda} \implies \lambda \in \mathbb{R} \]
Dependency for:
- Orthogonally diagonalizable iff hermitian
- Symmetric operator on V has a basis of orthonormal eigenvectors
Info:
- Depth: 12
- Number of transitive dependencies: 55
Transitive dependencies:
- /linear-algebra/vector-spaces/condition-for-subspace
- /linear-algebra/matrices/gauss-jordan-algo
- /complex-numbers/conjugation-is-homomorphic
- /sets-and-relations/composition-of-bijections-is-a-bijection
- /sets-and-relations/equivalence-relation
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- Vector space isomorphism is an equivalence relation
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- Inner product is anti-linear in second argument
- Symmetric operator
- A field is an integral domain
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