Definition 5

— Let E be a finite dimensional vector space over a commutative field K and let $\mathfrak{F}$ be a set of endomorphisms of E. Then $\mathfrak{F}$ is said to be diagonal with respect to a basis $(e_i)$ of E if the matrix of each $u \in \mathfrak{F}$ with respect to $(e_i)$ is diagonal. The set $\mathfrak{F}$ is said to be diagonalisable if there exists a basis of E with respect to which $\mathfrak{F}$ is diagonal.

This definition applies in particular to the case when $\mathfrak{F}$ contains only one element $u$; we then say that $u$ is diagonal (diagonalisable). Note also that $\mathfrak{F}$ is diagonal with respect to a basis $(e_i)$ if and only if the $(e_i)$ are common eigenvectors of all the elements of $\mathfrak{F}$; it follows that $\mathfrak{F}$ is diagonalisable if and only if E is generated by eigenvectors common to all the elements of $\mathfrak{F}$.

Let A be a subalgebra of $\mathrm{End}_K(E)$ containing Id. Then A is diagonalisable if and only if it is isomorphic to an algebra $K^r$ (in other words is diagonalisable in the sense of V, p. 28, Def. 1); indeed, if A is isomorphic to $K^r$, then A is diagonalisable by V, p. 29, Prop. 1; conversely, if A is diagonalisable, then it is isomorphic to a subalgebra of the algebra of diagonal matrices, which is isomorphic as an algebra to $K^n$, $n = \dim(E)$, hence A is isomorphic to some algebra $K^r$ (V, p. 30, Prop. 3).