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MatricesLinear algebraCollegeabout 1:02

Find the eigenvalues of the matrix [5 4; 1 2]: animated lesson

A narrated, step-by-step matrices video lesson with captions. Final answer: λ = 6, 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 [5 4; 1 2].
  2. 2. Steps, part 1
    1. det⁡(λI−A)=0,A=[5412]\det(\lambda I - A) = 0,\quad A = \begin{bmatrix}5 & 4 \\ 1 & 2\end{bmatrix}
      Set up the characteristic equation
    2. λ2−(7)λ+(6)=0\lambda^2 - (7)\lambda + (6) = 0
      For a 2×2 matrix it is λ² − (trace)λ + det
    3. p(λ)=λ2−7λ+6p(\lambda) = \lambda^{2} - 7\lambda + 6
      Characteristic polynomial
    4. λ=1, λ=6\lambda = 1,\ \lambda = 6
      Rational roots found with the rational root theorem
  3. 3. Steps, part 2
    1. [−141−4] v=0⇒v=⟨4,1⟩\begin{bmatrix}-1 & 4 \\ 1 & -4\end{bmatrix}\,v = 0 \Rightarrow v = \langle 4, 1 \rangle
      Eigenvectors for λ = 6: solve (A − λI)v = 0
    2. [4411] v=0⇒v=⟨−1,1⟩\begin{bmatrix}4 & 4 \\ 1 & 1\end{bmatrix}\,v = 0 \Rightarrow v = \langle -1, 1 \rangle
      Eigenvectors for λ = 1: solve (A − λI)v = 0
  4. 4. The answer
    λ=6, 1\lambda = 6,\ 1

    p(λ) roots: 6, 1

  5. 5. Check
    • Sum of eigenvalues ≈ 7 equals the trace 7.
    • Product of eigenvalues equals the determinant 6.

Transcript

The problem. Here is the problem: Find the eigenvalues of the matrix [5 4; 1 2]. 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 6: solve (A minus λI)v equals 0. Eigenvectors for λ equals 1: solve (A minus λI)v equals 0.

The answer. So the answer is λ equals 6, 1. The eigenvalues are the roots of the characteristic polynomial. Eigenvectors: λ equals 6: ⟨4, 1⟩; λ equals 1: ⟨-1, 1⟩.

Check. Sum of eigenvalues approximately 7 equals the trace 7. Product of eigenvalues equals the determinant 6.

About matrices

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

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Find the eigenvalues of the matrix [5 4; 1 2] (Lesson)