Why is it that the characteristic polynomial for a matrix $A$
$$\phi(\lambda) = \det(\lambda I – A)$$
when finding the roots gives the eigenvalues of $A$?
Why can the characteristic polynomial be used to find eigenvalues
eigenvalues-eigenvectorslinear algebrapolynomials
Best Answer
By definition, an eigenvalue of a matrix $A$ is a number $\lambda$ such that $Ax=\lambda x$ where $x$ is a non zero vector, called the eigenvector corresponding to $\lambda$. This means $(A-\lambda I)x=0$ for this $x$. That is, $\lambda$ is an eigenvalue of $A$ iff $A-\lambda I$ is not invertible. That is, $\lambda$ is an eigenvalue of $A$ iff $\det (A-\lambda I)=0$. Here $\det (A-\lambda I)$ is a polynomial in $\lambda$ and any $\lambda$'s satisfying this polynomial are eigenvalues!