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Linear-Algebra · Eigenvalues

Eigenvalues of [[-1,2],[2,-1]]

\left[\begin{matrix}-1 & 2\\2 & -1\end{matrix}\right]

Step by step

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

    Eigenvalues are the roots of the characteristic polynomial.

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

    Subtract λ from the diagonal.

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

    Expand the determinant.

  4. \left(\lambda - 1\right) \left(\lambda + 3\right) = 0

    Factor.

  5. \lambda = 1, \lambda = -3

    Eigenvalues (with multiplicity).

  6. \lambda = -3:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}-1\\1\end{matrix}\right]

    Solve (A − -3I)v = 0 for a basis eigenvector.

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

    Solve (A − 1I)v = 0 for a basis eigenvector.

Reveal the answer
\lambda = 1,\; \lambda = -3