maths.free › Linear Algebra › Eigenvalues and eigenvectors

Eigenvalues and eigenvectors

The characteristic polynomial and the directions a matrix only stretches.

An eigenvector is a direction a matrix does not rotate — only stretches by its eigenvalue. Find eigenvalues as roots of det(A − λI), then eigenvectors by solving (A − λI)v = 0. In the 3D view the eigenvectors are drawn on the transformed cube.

Worked example: eigenvalues of [[2,1],[1,2]]

Linear-Algebra · Eigenvalues

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

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

    Subtract λ from the diagonal.

  3. \lambda^{2} - 4 \lambda + 3 = 0

    Expand the determinant.

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

    Factor.

  5. \lambda = 3, \lambda = 1

    Eigenvalues (with multiplicity).

  6. \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.

  7. \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.

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

Try your own

More in Linear Algebra

DeterminantsMatrix inverseMatrix multiplicationRow reduction