Characteristic Polynomial¶

Definition¶

For a square matrix $\boldsymbol{A}$ , the characteristic polynomial is $\det(\boldsymbol{A}-\lambda \boldsymbol{I})$ $λ I)$ , viewed as a polynomial in $\lambda$ . The equation $\det(\boldsymbol{A}-\lambda \boldsymbol{I})=0$ is called the characteristic equation.

Interpretation¶

The characteristic polynomial turns the search for eigenvalues into a polynomial equation. A number $\lambda$ is an eigenvalue when $\boldsymbol{A}-\lambda\boldsymbol{I}$ sends some nonzero vector to $\boldsymbol{0}$ , and the determinant detects exactly this: it is zero when the matrix is singular.

The parabola lambda squared minus 4 lambda plus 3 crossing the horizontal axis at its roots 1 and 3, the eigenvalues of the matrix with rows (2, 1) and (1, 2).

For $\boldsymbol{A}=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ $(2112)$ , $\det(\boldsymbol{A}-\lambda\boldsymbol{I})=(2-\lambda)^2-1=\lambda^2-4\lambda+3$ $3$ . The parabola crosses the horizontal axis at $\lambda=1$ and $\lambda=3$ $3$ , and these roots are the eigenvalues of $\boldsymbol{A}$ , with eigenvectors $(1,-1)$ and $(1,1)$ .

Change the entries of $\boldsymbol{A}$ : the roots of $\det(\boldsymbol{A}-\lambda \boldsymbol{I})$ on the right are the eigenvalues, and their eigenvector lines are dashed on the left.

The identity matrix is needed because $\lambda$ is a number and $\boldsymbol{A}$ a matrix: $\lambda\boldsymbol{I}$ subtracts $\lambda$ from the diagonal entries only. Expanding the determinant gives a polynomial of degree $n$ ; for a $2\times2$ matrix it is $\lambda^2-\operatorname{tr}(\boldsymbol{A})\lambda+\det\boldsymbol{A}$ $det A$ , with the trace as a coefficient. Some texts use $\det(\lambda\boldsymbol{I}-\boldsymbol{A})$ instead, which differs by the factor $(-1)^n$ and has the same roots.

For a real matrix the polynomial has real coefficients, so the results on polynomials apply. By the fundamental theorem of algebra it has $n$ complex roots counted with multiplicity, and nonreal roots come in conjugate pairs. The quarter-turn rotation $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ has characteristic polynomial $\lambda^2+1$ and eigenvalues $\pm\mathrm{i}$ , but no real eigenvalue. The multiplicity of a root is the algebraic multiplicity of that eigenvalue.

A common slip is to subtract $\lambda$ from every entry, or to compute $\det\boldsymbol{A}-\lambda$ $λ$ . Only the diagonal is shifted, and the determinant is taken after the shift.

Key Results¶

The eigenvalues of $\boldsymbol{A}$ are exactly the roots of the characteristic polynomial, equivalently the solutions of the characteristic equation.
For an $n\times n$ matrix, the characteristic polynomial has degree $n$ .

Examples¶

For $\boldsymbol{A}=\begin{pmatrix}a&0\\0&b\end{pmatrix}$ $(a 0 0 b)$ , the characteristic polynomial is $(a-\lambda)(b-\lambda)$ $λ)$ .
For $\boldsymbol{I}_n$ , the characteristic polynomial is $(1-\lambda)^n$ $λ)^{n}$ .

Prerequisites¶

Sources¶

Math II: Linear Algebra · Week 6
Math V: Advanced Linear Algebra · Week 1 (revisited)

Keywords: characteristic equation Symbols: lambda

Last updated: 20 October 2026 03:02