Distinct Eigenvalues Theorem¶

Theorem¶

Fix a field $\mathbb{F}$ . If $\boldsymbol{A}$ is a linear operator with distinct eigenvalues $\lambda_1,\ldots,\lambda_n\in\mathbb{F}$ and corresponding eigenvectors $\boldsymbol{v}_1,\ldots,\boldsymbol{v}_n$ , then $(\boldsymbol{v}_1,\ldots,\boldsymbol{v}_n)$ is linearly independent.

Interpretation¶

Different eigenvalues produce genuinely different eigendirections. An eigenvector for $\lambda_1$ is scaled by $\lambda_1$ and one for $\lambda_2$ by a different factor, so neither can be built as a linear combination of eigenvectors with other eigenvalues.

For two eigenvectors the argument is short. Suppose $c_1\boldsymbol{v}_1+c_2\boldsymbol{v}_2=\boldsymbol{0}$ $0$ . Applying $\boldsymbol{A}$ gives $c_1\lambda_1\boldsymbol{v}_1+c_2\lambda_2\boldsymbol{v}_2=\boldsymbol{0}$ $0$ , and subtracting $\lambda_2$ times the first equation leaves $c_1(\lambda_1-\lambda_2)\boldsymbol{v}_1=\boldsymbol{0}$ $0$ . Since $\lambda_1\neq\lambda_2$ and the eigenvector $\boldsymbol{v}_1$ is nonzero, $c_1=0$ $0$ , and then $c_2=0$ $0$ . The general case repeats this step, removing one eigenvector at a time.

Distinctness is exactly what the argument uses. Two eigenvectors for the same eigenvalue can be dependent, such as $\boldsymbol{v}$ and $2\boldsymbol{v}$ , so the theorem takes one eigenvector per eigenvalue. It does not say that a repeated eigenvalue forces dependence: for $\boldsymbol{I}_2$ , the double eigenvalue $1$ has the independent eigenvectors $(1,0)$ and $(0,1)$ .

The main use is the first Key Result. An $n\times n$ matrix with $n$ distinct eigenvalues in $\mathbb{F}$ has $n$ independent eigenvectors, which form a basis of $\mathbb{F}^n$ , an eigenbasis, so the matrix is diagonalizable. For example, $\begin{pmatrix}2&1\\1&2\end{pmatrix}$ has the eigenvalues $3$ and $1$ , so its eigenvectors $(1,1)$ and $(1,-1)$ form a basis of $\mathbb{R}^2$ without further checking.

The condition is sufficient, not necessary. When an eigenvalue is repeated, the matrix may or may not be diagonalizable, and the algebraic and geometric multiplicities decide.

Key Results¶

An $n\times n$ matrix with $n$ distinct eigenvalues in $\mathbb{F}$ is diagonalizable over $\mathbb{F}$ .
The theorem is a main reason eigenvectors can form useful bases.

Examples¶

A $2\times2$ diagonal matrix with diagonal entries $2$ and $3$ has two independent eigenvectors.
A $3\times3$ matrix with three distinct eigenvalues has three independent eigenvectors.

Prerequisites¶

Sources¶

Math II: Linear Algebra · Week 6

Aliases: eigenvectors for distinct eigenvalues theorem Symbols: element of, lambda, real numbers

Last updated: 20 October 2026 03:02