---
tag: 053QVHF6N7
kind: statement
role: theorem
name: Eigenvalues are the roots of the characteristic polynomial
msc: [15A18]
uses: [0ZND7NQ970, 0BZ81AJNWF]
sources:
  - url: https://en.wikipedia.org/wiki/Characteristic_polynomial
    locator: Formal definition
---

## Statement

Let $F$ be a field, let $V$ be a finite-dimensional vector space over $F$, and let $T \colon V \to V$ be a linear map.
Let $p_T$ be the [[characteristic polynomial|0J5EMJRRFM]] of $T$, and let $\lambda \in F$.
Then $\lambda$ is an [[eigenvalue|0BQAA2BK3Q]] of $T$ if and only if $p_T(\lambda) = 0$.

## Remarks

The proof: $\lambda$ is an eigenvalue if and only if $\lambda \id_V - T$ is not injective, if and only if $\det(\lambda \id_V - T) = 0$.
So $T$ has at most $\dim V$ eigenvalues.
For $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, $p_A(t) = t^2 - 4t + 3 = (t - 1)(t - 3)$, so the eigenvalues are $1$ and $3$.

## Explanations

### L1 human

To find the eigenvalues of a matrix, compute its characteristic polynomial and find its roots.
Each root is an eigenvalue, and each eigenvalue is a root.
Then, for each eigenvalue $\lambda$, solve $(A - \lambda I)x = 0$ to find the eigenvectors.