math115: fix eigen def

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eggy 2022-12-04 18:24:38 -05:00
parent ae0489b522
commit 4812aafe94

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@ -943,8 +943,8 @@ $$(A-\lambda I)\vec x=\vec 0$$
The determinant of the system can be used to check if there will be any eigenvectors. The determinant of the system can be used to check if there will be any eigenvectors.
- If $\det(A-\lambda I)=0$, there is exactly one solution which is the trivial solution, so $\lambda$ is **not an eigenvalue**. - If $\det(A-\lambda I)=0$, there is exactly one solution which is the trivial solution, so $\lambda$ is **an eigenvalue**.
- If $\det(A-\lambda I)\neq 0$, there are multiple solutions, so $\lambda$ is an eigenvalue. - If $\det(A-\lambda I)\neq 0$, there are multiple solutions, so $\lambda$ is **not** an eigenvalue.
The **characteristic polynomial** of an eigenvalue is equal to its determinant, and can be used to solve for eigenvalues when $\lambda$ is unknown. The **characteristic polynomial** of an eigenvalue is equal to its determinant, and can be used to solve for eigenvalues when $\lambda$ is unknown.