Real matrix with real eigenvalues has real eigenvectors
Dependencies:
- Eigenvalues and Eigenvectors
- Conjugation of matrices is homomorphic
- /complex-numbers/complex-numbers
$\newcommand{\ubar}{\overline{u}}$ Let $A$ be a real matrix. Let $A$ have a real eigenvalue $\lambda$ and a corresponding complex eigenvector $u$. Then a real eigenvector of $A$ exists corresponding to the eigenvalue $\lambda$.
Proof
\begin{align} & Au = \lambda u \implies \overline{A}\,\ubar = \overline{\lambda}\,\ubar \implies A\ubar = \lambda\ubar \\ &\implies A(u + \ubar) = \lambda(u + \ubar) \textrm{ and } A(i(u - \ubar)) = \lambda(i(u - \ubar)) \end{align}
$u + \ubar = 2\operatorname{Re}(u)$ and $i(u - \ubar) = -2\operatorname{Im}(u)$ are real. At least one of them must be non-zero, since $u + \ubar = u - \ubar = 0 \implies u = 0$, which contradicts the fact that $u$ is an eigenvector of $A$. Hence, one of $u + \ubar$ and $i(u - \ubar)$ is a real eigenvector of $A$ corresponding to $\lambda$.
Dependency for:
- Orthogonally diagonalizable iff hermitian
- Symmetric operator on V has a basis of orthonormal eigenvectors
Info:
- Depth: 12
- Number of transitive dependencies: 52
Transitive dependencies:
- /linear-algebra/vector-spaces/condition-for-subspace
- /linear-algebra/matrices/gauss-jordan-algo
- /complex-numbers/conjugation-is-homomorphic
- /complex-numbers/complex-numbers
- /sets-and-relations/composition-of-bijections-is-a-bijection
- /sets-and-relations/equivalence-relation
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