Proposition 1

*Suppose $E$ is finite dimensional. Then an element $a$ of $K$ is an eigenvalue of the endomorphism $u$ if and only if it is a root of the characteristic polynomial of $u$.

If $L$ is an extension of the field $K$, then the roots of $\chi_u$ in $L$ are eigenvalues of the endomorphism $1, \otimes u$ of the $L$-vector space $L \otimes_K E$. They are often referred to as eigenvalues of $u$ in $L$. By abuse of language, we say that all the eigenvalues of $u$ belong to $L$ if this holds for all the eigenvalues of $u$ in an algebraically closed extension of $L$; this means that $\chi_u$ decomposes into linear factors in $L[X]$.

Let $U$ be a square matrix of order $n$ with coefficients in $K$. Then by definition the characteristic polynomial of $U$ is

the eigenvalues of $U$ (in an extension $L$ of $K$) are the roots (in $L$) of the polynomial $\chi_U$; these are also the scalars $a$ (in $L$) such that there exists a nonzero solution to the system of linear equations $UX = \alpha X$, where $X$ is a column matrix of order $n$; a column matrix $X$ satisfying this equation is called an eigenvector of $U$ corresponding to $a$.

If $U$ is the matrix of an endomorphism $u$ of an n-dimensional vector space with respect to a basis $B$, then $\chi_U = \chi_u$, the eigenvalues of $U$ are the eigenvalues of $u$, and the eigenvectors of $U$ are the matrices of the eigenvectors of $u$ with respect to the basis $B$.