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MatricesLinear algebraCollegeabout 1:01

Find the eigenvalues of the matrix [6 -2; 2 1]: animated lesson

A narrated, step-by-step matrices video lesson with captions. Final answer: λ = 5, 2.

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 [6 -2; 2 1].
  2. 2. Steps, part 1
    1. det⁡(λI−A)=0,A=[6−221]\det(\lambda I - A) = 0,\quad A = \begin{bmatrix}6 & -2 \\ 2 & 1\end{bmatrix}
      Set up the characteristic equation
    2. λ2−(7)λ+(10)=0\lambda^2 - (7)\lambda + (10) = 0
      For a 2×2 matrix it is λ² − (trace)λ + det
    3. p(λ)=λ2−7λ+10p(\lambda) = \lambda^{2} - 7\lambda + 10
      Characteristic polynomial
    4. λ=2, λ=5\lambda = 2,\ \lambda = 5
      Rational roots found with the rational root theorem
  3. 3. Steps, part 2
    1. [1−22−4] v=0⇒v=⟨2,1⟩\begin{bmatrix}1 & -2 \\ 2 & -4\end{bmatrix}\,v = 0 \Rightarrow v = \langle 2, 1 \rangle
      Eigenvectors for λ = 5: solve (A − λI)v = 0
    2. [4−22−1] v=0⇒v=⟨1,2⟩\begin{bmatrix}4 & -2 \\ 2 & -1\end{bmatrix}\,v = 0 \Rightarrow v = \langle 1, 2 \rangle
      Eigenvectors for λ = 2: solve (A − λI)v = 0
  4. 4. The answer
    λ=5, 2\lambda = 5,\ 2

    p(λ) roots: 5, 2

  5. 5. Check
    • Sum of eigenvalues ≈ 7 equals the trace 7.
    • Product of eigenvalues equals the determinant 10.

Transcript

The problem. Here is the problem: Find the eigenvalues of the matrix [6 minus 2; 2 1]. Find the eigenvalues and eigenvectors of a 2 times 2 matrix.

Steps, part 1. Set up the characteristic equation. For a 2 times 2 matrix it is λ squared minus (trace)λ plus det. Characteristic polynomial. Rational roots found with the rational root theorem.

Steps, part 2. Eigenvectors for λ equals 5: solve (A minus λI)v equals 0. Eigenvectors for λ equals 2: solve (A minus λI)v equals 0.

The answer. So the answer is λ equals 5, 2. The eigenvalues are the roots of the characteristic polynomial. Eigenvectors: λ equals 5: ⟨2, 1⟩; λ equals 2: ⟨1, 2⟩.

Check. Sum of eigenvalues approximately 7 equals the trace 7. Product of eigenvalues equals the determinant 10.

About matrices

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

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Find the eigenvalues of the matrix [6 -2; 2 1] (Lesson)