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Module 3/Eigenvalues & Diagonalization

Eigenvalues, Eigenvectors & the Characteristic Polynomial

Meet the equation Av=λvA\mathbf{v}=\lambda\mathbf{v}. This lesson explains what eigenvalues and eigenvectors are, why the eigenspace EλE_\lambda is exactly the kernel of A−λIA-\lambda I, and how the characteristic polynomial cA(λ)=det⁡(A−λI)c_A(\lambda)=\det(A-\lambda I) turns the search for eigenvalues into solving a single polynomial equation.

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

Which of these vectors is an eigenvector of A=[3113]A=\begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix} for the eigenvalue λ=4\lambda=4?

For A=[0−1−1121112]A=\begin{bmatrix} 0 & -1 & -1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}, the eigenvalue λ=1\lambda=1 has eigenspace E1=ker⁡(A−I)E_1=\ker(A-I). What is dim⁡(E1)\dim(E_1)?

What you’ll be able to do

  • State Definition 7.1: a scalar λ\lambda is an eigenvalue of a square matrix AA when Av=λvA\mathbf{v}=\lambda\mathbf{v} for some nonzero vector v\mathbf{v}, and explain (Remark 7.1) why v≠0\mathbf{v}\neq\mathbf{0} is required.
  • Describe the eigenspace Eλ={v:Av=λv}E_\lambda=\{\mathbf{v}:A\mathbf{v}=\lambda\mathbf{v}\} and use Theorem 7.1 to identify it with ker⁡(A−λI)\ker(A-\lambda I).
  • Set up and solve the characteristic equation det⁡(A−λI)=0\det(A-\lambda I)=0 (Corollary 7.1) to find all eigenvalues of AA.
  • Build the characteristic polynomial cA(λ)=det⁡(A−λI)c_A(\lambda)=\det(A-\lambda I) (Theorem 7.2), and for 2×22\times 2 matrices use cA(λ)=λ2−tr⁡(A) λ+det⁡(A)c_A(\lambda)=\lambda^2-\operatorname{tr}(A)\,\lambda+\det(A) together with the discriminant to count real eigenvalues.
  • Find eigenvectors and the dimension of each eigenspace by solving the homogeneous system (A−λI)v=0(A-\lambda I)\mathbf{v}=\mathbf{0}.

In your course

· MATH2015 · Linear Algebra & Probability
§7.1 Eigenvalues and eigenvectors
  • Definition 7.1Eigenvalue and eigenvector
    For an n×nn\times n matrix AA, a scalar λ\lambda is an eigenvalue if Av=λvA\mathbf{v}=\lambda\mathbf{v} for some nonzero v\mathbf{v}; such a v\mathbf{v} is an eigenvector corresponding to λ\lambda.
  • Remark 7.1Why the eigenvector must be nonzero
    v=0\mathbf{v}=\mathbf{0} solves Av=λvA\mathbf{v}=\lambda\mathbf{v} for every λ\lambda, so it is excluded as a trivial solution.
  • Definition 7.2Eigenspace
    Eλ={v∈Rn:Av=λv}E_\lambda=\{\mathbf{v}\in\mathbb{R}^n:A\mathbf{v}=\lambda\mathbf{v}\}, which also contains the zero vector.
  • Theorem 7.1Eigenvalues and nontrivial kernels
    λ\lambda is an eigenvalue of AA iff rank⁡(A−λI)<n\operatorname{rank}(A-\lambda I)<n iff A−λIA-\lambda I has a nontrivial kernel; moreover Eλ=ker⁡(A−λI)E_\lambda=\ker(A-\lambda I).
  • Corollary 7.1Characteristic equation
    λ\lambda is an eigenvalue of AA iff det⁡(A−λI)=0\det(A-\lambda I)=0.
  • Theorem 7.2Characteristic polynomial
    cA(λ)=det⁡(A−λIn)c_A(\lambda)=\det(A-\lambda I_n) has degree nn: cA(λ)=(−1)nλn+(−1)n−1tr⁡(A)λn−1+⋯+det⁡(A)c_A(\lambda)=(-1)^n\lambda^n+(-1)^{n-1}\operatorname{tr}(A)\lambda^{n-1}+\cdots+\det(A).
  • Corollary 7.2Eigenvalues, determinant, and trace
    For eigenvalues λ1,…,λn\lambda_1,\dots,\lambda_n (with multiplicity), det⁡A=λ1⋯λn\det A=\lambda_1\cdots\lambda_n and tr⁡A=λ1+⋯+λn\operatorname{tr} A=\lambda_1+\cdots+\lambda_n.
Worked examples follow Examples 7.1-7.3 in the notes (including the 2×22\times 2 discriminant cases and the 3×33\times 3 matrix of Example 7.3). Algebraic multiplicity (Definition 7.3, Theorem 7.3) and complex eigenvalues of real matrices (Example 7.5, Proposition 7.1) are introduced in this section but are developed further in later lessons on diagonalization; this lesson concentrates on the real-eigenvalue mechanics of the characteristic polynomial.
1

What an eigenvalue and eigenvector are

Let AA be an n×nn\times n matrix. A scalar λ\lambda is an eigenvalue of AA if there is a nonzero vector v≠0\mathbf{v}\neq\mathbf{0} with Av=λv.A\mathbf{v}=\lambda\mathbf{v}. Such a v\mathbf{v} is an eigenvector corresponding to λ\lambda (Definition 7.1). Geometrically, AA represents a linear transformation, and an eigenvector points in a direction that is left unchanged by AA: the vector is only stretched or compressed by the factor λ\lambda. If λ<0\lambda<0 the direction is reversed as well. The requirement v≠0\mathbf{v}\neq\mathbf{0} matters (Remark 7.1): v=0\mathbf{v}=\mathbf{0} satisfies Av=λvA\mathbf{v}=\lambda\mathbf{v} for every scalar λ\lambda, so it is a trivial solution that would tell us nothing.

2

Eigenspaces and the kernel of $A-\lambda I$

For a fixed eigenvalue λ\lambda, the set of all vectors satisfying Av=λvA\mathbf{v}=\lambda\mathbf{v} is the eigenspace Eλ={v∈Rn:Av=λv}E_\lambda=\{\mathbf{v}\in\mathbb{R}^n : A\mathbf{v}=\lambda\mathbf{v}\} (Definition 7.2); it includes the zero vector. Rewriting the eigenvalue equation, Av=λv  ⟺  Av−λIv=0  ⟺  (A−λI)v=0.A\mathbf{v}=\lambda\mathbf{v}\iff A\mathbf{v}-\lambda I\mathbf{v}=\mathbf{0}\iff (A-\lambda I)\mathbf{v}=\mathbf{0}. So EλE_\lambda is precisely the kernel of A−λIA-\lambda I. A homogeneous system has a nonzero solution exactly when its coefficient matrix has a nontrivial kernel, which is the key idea behind Theorem 7.1: Eλ=ker⁡(A−λI)E_\lambda=\ker(A-\lambda I), and the eigenvectors for λ\lambda are the nonzero elements of this kernel.

3

Finding eigenvalues: the characteristic equation

By Theorem 7.1, λ\lambda is an eigenvalue iff A−λIA-\lambda I has a nontrivial kernel, i.e. rank⁡(A−λI)<n\operatorname{rank}(A-\lambda I)<n, i.e. A−λIA-\lambda I is singular. A square matrix is singular exactly when its determinant is zero, which gives Corollary 7.1: λ is an eigenvalue of A  ⟺  det⁡(A−λI)=0.\lambda\text{ is an eigenvalue of }A\iff \det(A-\lambda I)=0. This is the characteristic equation. The practical two-step recipe is: (1) solve det⁡(A−λI)=0\det(A-\lambda I)=0 to get all eigenvalues; (2) for each eigenvalue λ\lambda, solve (A−λI)v=0(A-\lambda I)\mathbf{v}=\mathbf{0} (e.g. by Gaussian elimination) to describe the eigenspace Eλ=ker⁡(A−λI)E_\lambda=\ker(A-\lambda I) and read off its eigenvectors.

4

The characteristic polynomial and the $2\times 2$ formula

The function cA(λ)=det⁡(A−λI)c_A(\lambda)=\det(A-\lambda I) is the characteristic polynomial of AA; by Theorem 7.2 it has degree nn, with form cA(λ)=(−1)nλn+(−1)n−1tr⁡(A) λn−1+⋯+det⁡(A),c_A(\lambda)=(-1)^n\lambda^n+(-1)^{n-1}\operatorname{tr}(A)\,\lambda^{n-1}+\cdots+\det(A), so the trace controls the second-highest coefficient and the constant term is det⁡(A)\det(A). For a 2×22\times 2 matrix A=[abcd]A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} this becomes the handy formula cA(λ)=λ2−tr⁡(A) λ+det⁡(A),c_A(\lambda)=\lambda^2-\operatorname{tr}(A)\,\lambda+\det(A), a quadratic whose roots are λ1,2=tr⁡(A)±tr⁡(A)2−4det⁡(A)2\lambda_{1,2}=\dfrac{\operatorname{tr}(A)\pm\sqrt{\operatorname{tr}(A)^2-4\det(A)}}{2}. The discriminant tr⁡(A)2−4det⁡(A)\operatorname{tr}(A)^2-4\det(A) decides the count: negative gives 00 real eigenvalues, zero gives 11 (a repeated eigenvalue), and positive gives 22 distinct real eigenvalues.

Theorem 7.1 (Eigenvalues and nontrivial kernels)

A scalar λ\lambda is an eigenvalue of the n×nn\times n matrix AA if and only if rank⁡(A−λI)<n\operatorname{rank}(A-\lambda I)<n, equivalently if and only if A−λIA-\lambda I has a nontrivial kernel. The eigenvectors for λ\lambda are the nonzero solutions of (A−λI)v=0(A-\lambda I)\mathbf{v}=\mathbf{0}, and the eigenspace is Eλ=ker⁡(A−λI)E_\lambda=\ker(A-\lambda I).

Intuition. Av=λvA\mathbf{v}=\lambda\mathbf{v} rearranges to (A−λI)v=0(A-\lambda I)\mathbf{v}=\mathbf{0}. Wanting a nonzero solution is the same as asking the matrix A−λIA-\lambda I to collapse some direction to zero, i.e. to have a nontrivial kernel and therefore deficient rank.
Corollary 7.1 (Characteristic equation)

A scalar λ\lambda is an eigenvalue of AA if and only if det⁡(A−λI)=0\det(A-\lambda I)=0. This is the characteristic equation of AA.

Intuition. A matrix has a nontrivial kernel exactly when it is singular, and singular is detected by a zero determinant. So the kernel condition of Theorem 7.1 becomes a single scalar equation you can solve.
Theorem 7.2 (Characteristic polynomial)

If AA is n×nn\times n, then cA(λ)=det⁡(A−λIn)c_A(\lambda)=\det(A-\lambda I_n) is a polynomial of degree nn: cA(λ)=(−1)nλn+(−1)n−1tr⁡(A) λn−1+⋯+det⁡(A)c_A(\lambda)=(-1)^n\lambda^n+(-1)^{n-1}\operatorname{tr}(A)\,\lambda^{n-1}+\cdots+\det(A). In particular, for a 2×22\times 2 matrix cA(λ)=λ2−tr⁡(A) λ+det⁡(A)c_A(\lambda)=\lambda^2-\operatorname{tr}(A)\,\lambda+\det(A).

Intuition. Expanding det⁡(A−λI)\det(A-\lambda I), the diagonal product (a11−λ)⋯(ann−λ)(a_{11}-\lambda)\cdots(a_{nn}-\lambda) supplies the top-degree terms (−λ)n+tr⁡(A)(−λ)n−1(-\lambda)^n+\operatorname{tr}(A)(-\lambda)^{n-1}; every other term of the determinant uses at least two off-diagonal entries and so has lower degree. Setting λ=0\lambda=0 shows the constant term is det⁡(A)\det(A).
Corollary 7.2 (Eigenvalues, determinant, and trace)

If an n×nn\times n matrix AA has eigenvalues λ1,λ2,…,λn\lambda_1,\lambda_2,\dots,\lambda_n listed with their algebraic multiplicities, then det⁡A=λ1λ2⋯λn\det A=\lambda_1\lambda_2\cdots\lambda_n (the product of the eigenvalues) and tr⁡A=λ1+λ2+⋯+λn\operatorname{tr} A=\lambda_1+\lambda_2+\cdots+\lambda_n (their sum).

Intuition. The eigenvalues are the roots of cA(λ)c_A(\lambda), so comparing the factored form with the coefficients from Theorem 7.2 ties the product of the roots to the constant term det⁡(A)\det(A) and the sum of the roots to the trace. It is a fast sanity check on any eigenvalue computation.

Worked examples

Example 1

Find all eigenvalues and eigenspaces of A=[2112]A=\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}.

  1. 1

    Form A−λI=[2−λ112−λ]A-\lambda I=\begin{bmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{bmatrix}.

  2. 2

    Characteristic equation (Corollary 7.1): det⁡(A−λI)=(2−λ)2−1=0\det(A-\lambda I)=(2-\lambda)^2-1=0. Expanding gives λ2−4λ+3=0\lambda^2-4\lambda+3=0, which matches λ2−tr⁡(A)λ+det⁡(A)\lambda^2-\operatorname{tr}(A)\lambda+\det(A) with tr⁡(A)=4, det⁡(A)=3\operatorname{tr}(A)=4,\ \det(A)=3.

  3. 3

    Factor: λ2−4λ+3=(λ−1)(λ−3)=0\lambda^2-4\lambda+3=(\lambda-1)(\lambda-3)=0, so λ1=1\lambda_1=1 and λ2=3\lambda_2=3.

  4. 4

    For λ=3\lambda=3: A−3I=[−111−1]A-3I=\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} gives −v1+v2=0-v_1+v_2=0, so v1=v2v_1=v_2 and E3=span⁡{[11]}E_3=\operatorname{span}\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}\right\}.

  5. 5

    For λ=1\lambda=1: A−I=[1111]A-I=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} gives v1+v2=0v_1+v_2=0, so E1=span⁡{[1−1]}E_1=\operatorname{span}\left\{\begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}.

  6. 6

    Check: A[11]=[33]=3[11]A\begin{bmatrix} 1 \\ 1 \end{bmatrix}=\begin{bmatrix} 3 \\ 3 \end{bmatrix}=3\begin{bmatrix} 1 \\ 1 \end{bmatrix} and A[1−1]=[1−1]=1[1−1]A\begin{bmatrix} 1 \\ -1 \end{bmatrix}=\begin{bmatrix} 1 \\ -1 \end{bmatrix}=1\begin{bmatrix} 1 \\ -1 \end{bmatrix}.

Answer. Eigenvalues λ1=1\lambda_1=1 with E1=span⁡{[1−1]}E_1=\operatorname{span}\left\{\begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\} and λ2=3\lambda_2=3 with E3=span⁡{[11]}E_3=\operatorname{span}\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}\right\}.
Example 2

Use the 2×22\times 2 characteristic-polynomial formula to find the eigenvalues of A=[1221]A=\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}, and say how many real eigenvalues it has.

Example 3

Find the eigenvalues and eigenspaces of A=[0−1−1121112]A=\begin{bmatrix} 0 & -1 & -1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} (Example 7.3).