Solve any maths problem

Equations, derivatives, integrals, matrices, triangles, primes, statistics, or a word problem the tutor breaks into parts.

Linear-Algebra · Eigenvalues

Eigenvalues of [[2,1,0],[1,2,0],[0,0,1]]

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

    Subtract λ from the diagonal.

  3. - \lambda^{3} + 5 \lambda^{2} - 7 \lambda + 3 = 0

    Expand the determinant.

  4. - \left(\lambda - 3\right) \left(\lambda - 1\right)^{2} = 0

    Factor.

  5. \lambda = 3, \lambda = 1\ (\times 2)

    Eigenvalues (with multiplicity).

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

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

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

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

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

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

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