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Linear-Algebra · Eigenvalues

Eigenvalues of [[1,1],[1,0]]

\left[\begin{matrix}1 & 1\\1 & 0\end{matrix}\right]

Hakbang-hakbang

  1. \det(A - \lambda I) = 0

    Eigenvalues are the roots of the characteristic polynomial.

  2. \det\left[\begin{matrix}1 - \lambda & 1\\1 & - \lambda\end{matrix}\right] = 0

    Subtract λ from the diagonal.

  3. \lambda^{2} - \lambda - 1 = 0

    Expand the determinant.

  4. \lambda = \frac{1}{2} - \frac{\sqrt{5}}{2}, \lambda = \frac{1}{2} + \frac{\sqrt{5}}{2}

    Eigenvalues (with multiplicity).

  5. \lambda = \frac{1}{2} - \frac{\sqrt{5}}{2}:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}\frac{1}{2} - \frac{\sqrt{5}}{2}\\1\end{matrix}\right]

    Solve (A − \frac{1}{2} - \frac{\sqrt{5}}{2}I)v = 0 for a basis eigenvector.

  6. \lambda = \frac{1}{2} + \frac{\sqrt{5}}{2}:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}\frac{1}{2} + \frac{\sqrt{5}}{2}\\1\end{matrix}\right]

    Solve (A − \frac{1}{2} + \frac{\sqrt{5}}{2}I)v = 0 for a basis eigenvector.

Ipahayag ang sagot
\lambda = \frac{1}{2} - \frac{\sqrt{5}}{2},\; \lambda = \frac{1}{2} + \frac{\sqrt{5}}{2}