Eigenvalue-Root Theorem¶

Theorem¶

For any square matrix $\boldsymbol{A}$ with entries in a field $\mathbb{F}$, the eigenvalues of $\boldsymbol{A}$ are exactly the roots in $\mathbb{F}$ of its characteristic polynomial $\det(\boldsymbol{A}-\lambda \boldsymbol{I})$$\lambda \mathbf{I})$.

Interpretation¶

The theorem makes eigenvalues computable. The definition $\boldsymbol{A}\boldsymbol{p}=\lambda\boldsymbol{p}$ involves the unknown vector $\boldsymbol{p}$ as well as $\lambda$; the theorem removes $\boldsymbol{p}$ and leaves one equation in $\lambda$ alone, $\det(\boldsymbol{A}-\lambda\boldsymbol{I})=0$$0$.

The proof is a chain of equivalences. Using the identity matrix, $\boldsymbol{A}\boldsymbol{p}=\lambda\boldsymbol{p}$ becomes $(\boldsymbol{A}-\lambda\boldsymbol{I})\boldsymbol{p}=\boldsymbol{0}$$0$. So $\lambda$ is an eigenvalue if and only if this system has a solution $\boldsymbol{p}\neq\boldsymbol{0}$$0$, if and only if the kernel of $\boldsymbol{A}-\lambda\boldsymbol{I}$ is not $\{\boldsymbol{0}\}$, if and only if this square matrix is singular, if and only if its determinant is $0$.

The phrase "roots in $\mathbb{F}$" matters. For $\begin{pmatrix}1&-2\\2&1\end{pmatrix}$ the characteristic polynomial is $\lambda^2-2\lambda+5$$5$, with roots $1\pm2\mathrm{i}$$2\mathrm{i}$. Over $\mathbb{C}$ these are eigenvalues; over $\mathbb{R}$ the matrix has none, since no real $\lambda$ makes $\boldsymbol{A}-\lambda\boldsymbol{I}$ singular.

In practice the theorem splits the eigenvalue problem into two steps: first find the roots of the characteristic polynomial, then, for each root, solve the homogeneous system $(\boldsymbol{A}-\lambda\boldsymbol{I})\boldsymbol{p}=\boldsymbol{0}$$0$, for instance by Gauss-Jordan elimination, to find the eigenvectors. The number of times a root is repeated is its algebraic multiplicity.

The determinant step gives only $\lambda$, not the eigenvectors. And the homogeneous system always has the solution $\boldsymbol{p}=\boldsymbol{0}$$0$, which does not count: if elimination finds no other solution, the value of $\lambda$ was not a root after all.

Key Results¶

• $\lambda$ is an eigenvalue if and only if $\det(\boldsymbol{A}-\lambda \boldsymbol{I})=0$$0$.
• Every eigenvalue has at least one corresponding eigenvector.

Examples¶

• For a diagonal matrix, the roots are precisely the diagonal entries.
• For $\boldsymbol{A}=\begin{pmatrix}1&-2\\2&1\end{pmatrix}$$\left(\begin{array}{cc}1& -2\\ 2& 1\end{array}\right)$, the eigenvalues are $1\pm2\mathrm{i}$$2\mathrm{i}$.

Prerequisites¶

• Eigenvalue and Eigenvector
• Characteristic Polynomial
• Determinant

Related Concepts¶

• Root of a Polynomial
• Algebraic and Geometric Multiplicity

Sources¶

• Math II: Linear Algebra · Week 6

Keywords: eigenvalue theorem, characteristic root theorem Symbols: complex numbers, lambda, real numbers

Last updated: 20 October 2026 03:02