Characteristic Polynomial¶

Definition¶

For a square matrix $\boldsymbol{A}$, the characteristic polynomial is $\det(\boldsymbol{A}-\lambda \boldsymbol{I})$$\lambda \mathbf{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.

For $\boldsymbol{A}=\begin{pmatrix}2&1\\1&2\end{pmatrix}$$\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, $\det(\boldsymbol{A}-\lambda\boldsymbol{I})=(2-\lambda)^2-1=\lambda^2-4\lambda+3$$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 $(1,1)$.

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 \mathbf{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$$\lambda$. 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}$$\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)$, the characteristic polynomial is $(a-\lambda)(b-\lambda)$$\lambda )$.
• For $\boldsymbol{I}_n$, the characteristic polynomial is $(1-\lambda)^n$$\lambda {)}^n$.

Prerequisites¶

• Matrix
• Determinant
• Identity Matrix
• Polynomial

Related Concepts¶

• Root of a Polynomial
• Diagonalization
• Eigenvalue and Eigenvector
• Complex Conjugate Root Theorem
• Distinct Eigenvalues Theorem

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