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MatricesLinear algebraCollegeabout 1:07

Find the eigenvalues of the matrix [2 1 0; 1 2 0; 0 0 3]: animated lesson

A narrated, step-by-step matrices video lesson with captions. Final answer: λ = 3, 3, 1.

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.4 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    Find the eigenvalues of the matrix [2 1 0; 1 2 0; 0 0 3].
  2. 2. Steps, part 1
    1. det⁡(λI−A)=0,A=[210120003]\det(\lambda I - A) = 0,\quad A = \begin{bmatrix}2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3\end{bmatrix}
      Set up the characteristic equation
    2. p(λ)=λ3−7λ2+15λ−9p(\lambda) = \lambda^{3} - 7\lambda^{2} + 15\lambda - 9
      Characteristic polynomial
    3. λ=1, λ=3, λ=3\lambda = 1,\ \lambda = 3,\ \lambda = 3
      Rational roots found with the rational root theorem
    4. [−1101−10000] v=0⇒v=⟨1,1,0⟩, ⟨0,0,1⟩\begin{bmatrix}-1 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 0\end{bmatrix}\,v = 0 \Rightarrow v = \langle 1, 1, 0 \rangle,\ \langle 0, 0, 1 \rangle
      Eigenvectors for λ = 3: solve (A − λI)v = 0
  3. 3. Steps, part 2
    1. [110110002] v=0⇒v=⟨−1,1,0⟩\begin{bmatrix}1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 2\end{bmatrix}\,v = 0 \Rightarrow v = \langle -1, 1, 0 \rangle
      Eigenvectors for λ = 1: solve (A − λI)v = 0
  4. 4. The answer
    λ=3, 3, 1\lambda = 3,\ 3,\ 1

    p(λ) roots: 3, 3, 1

  5. 5. Check
    • Sum of eigenvalues ≈ 7 equals the trace 7.
    • Product of eigenvalues equals the determinant 9.

Transcript

The problem. Here is the problem: Find the eigenvalues of the matrix [2 1 0; 1 2 0; 0 0 3]. Find the eigenvalues and eigenvectors of a 3 times 3 matrix.

Steps, part 1. Set up the characteristic equation. Characteristic polynomial. Rational roots found with the rational root theorem. Eigenvectors for λ equals 3: solve (A minus λI)v equals 0. Algebraic multiplicity 2, geometric multiplicity 2.

Steps, part 2. Eigenvectors for λ equals 1: solve (A minus λI)v equals 0.

The answer. So the answer is λ equals 3, 3, 1. The eigenvalues are the roots of the characteristic polynomial. Eigenvectors: λ equals 3: ⟨1, 1, 0⟩, ⟨0, 0, 1⟩; λ equals 1: ⟨-1, 1, 0⟩.

Check. Sum of eigenvalues approximately 7 equals the trace 7. Product of eigenvalues equals the determinant 9.

About matrices

Determinants, inverses, products, transposes and eigenvalues of matrices.

More lessons like this

Find the eigenvalues of the matrix [2 1 0; 1 2 0; 0 0 3]