HexMathv1.0.3

Eigenvalues and eigenvectors calculator

Type or scan a square matrix. HexMath solves its characteristic equation for the eigenvalues, step by step.

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What this page covers

  • Eigenvalues of 2×2 and 3×3 matrices from the characteristic equation det⁡(A−λI)=0\det(A - \lambda I) = 0.
  • An eigenvector for each eigenvalue, by solving (A−λI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0}.
  • Triangular and block matrices, where the eigenvalues can be read off quickly.
  • Not covered here: the characteristic polynomial on its own (use Characteristic polynomial calculator) and determinants in general (use Determinant calculator).

How to enter the problem

  • Type it or paste it. Write eigenvalues of, then the matrix row by row, for example [[4, 1], [2, 3]], or use the math keyboard's matrix key.
  • Scan it. Fit the one problem in the frame.
  • Choose a photo. Several problems in one photo are solved as a list.
  • Then tap Solve.

Worked examples

Example 1

Problem

Find the eigenvalues and eigenvectors of (2112)\displaystyle \begin{pmatrix}2 & 1\\ 1 & 2\end{pmatrix}.

Answer

Verified

The answer is

λ=1,λ=3\lambda = 1, \lambda = 3

Explanation

  1. Set up the characteristic equation
    det⁡(2−λ112−λ)=(2−λ)2−1=0\det\begin{pmatrix}2 - \lambda & 1\\ 1 & 2 - \lambda\end{pmatrix} = (2 - \lambda)^2 - 1 = 0
  2. Solve it
    2−λ=±12 - \lambda = \pm 1, so λ=1\lambda = 1 or λ=3\lambda = 3.
  3. Eigenvector for λ=3\lambda = 3
    (A−3I)v=0(A - 3I)\mathbf{v} = \mathbf{0} gives −v1+v2=0-v_1 + v_2 = 0, so v=(1,1)\mathbf{v} = (1, 1).
  4. Eigenvector for λ=1\lambda = 1
    (A−I)v=0(A - I)\mathbf{v} = \mathbf{0} gives v1+v2=0v_1 + v_2 = 0, so v=(1,−1)\mathbf{v} = (1, -1).

Example 2

Problem

Find the eigenvalues of (500012021)\displaystyle \begin{pmatrix}5 & 0 & 0\\ 0 & 1 & 2\\ 0 & 2 & 1\end{pmatrix}.

Answer

Verified

The answer is

λ=−1,λ=3,λ=5\lambda = -1, \lambda = 3, \lambda = 5

Explanation

  1. Expand along the first row
    det⁡(A−λI)=(5−λ)[(1−λ)2−4]\det(A - \lambda I) = (5 - \lambda)\left[(1 - \lambda)^2 - 4\right], because the rest of the first row is zero.
  2. Set each factor to zero
    5−λ=05 - \lambda = 0 gives λ=5\lambda = 5. (1−λ)2=4(1 - \lambda)^2 = 4 gives 1−λ=±21 - \lambda = \pm 2.
  3. Solve the second factor
    λ=−1\lambda = -1 or λ=3\lambda = 3.

Example 3

Problem

Find the eigenvalues of the upper triangular matrix (34−10−25001)\displaystyle \begin{pmatrix}3 & 4 & -1\\ 0 & -2 & 5\\ 0 & 0 & 1\end{pmatrix}.

Answer

Verified

The answer is

λ=−2,λ=1,λ=3\lambda = -2, \lambda = 1, \lambda = 3

Explanation

  1. Use the triangular shape
    A−λIA - \lambda I is still triangular, and the determinant of a triangular matrix is the product of its diagonal.
  2. Write the characteristic equation
    (3−λ)(−2−λ)(1−λ)=0(3 - \lambda)(-2 - \lambda)(1 - \lambda) = 0
  3. Read off the roots
    The eigenvalues are the diagonal entries 33, −2-2 and 11.

Common mistakes

  • Reading the diagonal entries as eigenvalues when the matrix is not triangular. For the runnable example, 44 and 33 are not eigenvalues; 22 and 55 are.
  • Subtracting λ\lambda from every entry instead of only the diagonal entries.
  • Sign errors in the characteristic polynomial. For a 2×2 matrix it is λ2−(trace)λ+det⁡A\lambda^2 - (\text{trace})\lambda + \det A.
  • Giving the zero vector as an eigenvector. An eigenvector must be nonzero.

Checks, assumptions and limits

  • Check with the trace and the determinant. The eigenvalues add up to the trace and multiply to the determinant. For the runnable example, 2+5=7=4+32 + 5 = 7 = 4 + 3 and 2⋅5=10=4⋅3−1⋅22 \cdot 5 = 10 = 4 \cdot 3 - 1 \cdot 2.
  • Eigenvalues exist only for square matrices. Some real matrices have complex eigenvalues; this page works with real ones, and eigenvectors are shown up to a nonzero multiple.
  • HexMath can make mistakes. Double check important steps.

Related tasks

Frequently asked questions

Yes, when you ask for them. For each eigenvalue the steps solve (A − λI)v = 0 and give one eigenvector. Any nonzero multiple of it is also an eigenvector.

Yes. The steps expand the determinant of A − λI into a cubic equation and solve it, using a zero row or a triangular shape when there is one.

Write eigenvalues of, then the matrix row by row in square brackets, such as [[4, 1], [2, 3]]. Or scan the problem with the camera.

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