Proposition 12

— Let E be a finite dimensional vector space over a commutative field K, and let u be an endomorphism of E. Then the following conditions are equivalent:

(i) u is diagonalisable. (ii) E is the direct sum of the eigenspaces of u. (iii) All the roots of the minimal polynomial of u are in K, and these roots are all simple.

Moreover, if these conditions are satisfied, then every subspace of E closed under u is the direct sum of its intersections with the eigenspaces of u.

The equivalence of (i) and (ii) follows from the preceding remarks and VII, p. 31, Prop. 2. Suppose u is diagonalisable, and let $(\alpha_i)$ be its family of eigenvalues, and $(V_i)$ the corresponding family of eigenspaces; since the restriction of u to $V_i$ is the homothety defined by $\alpha_i$, it annihilates the polynomial $X - \alpha_i$; it follows that u annihilates the polynomial $\prod (X - \alpha_i)$, which is therefore a multiple of the minimal polynomial of u, and hence coincides with it, which proves (iii). Conversely, if (iii) is satisfied then there exists a basis of E with respect to which the matrix U of u is a diagonal block of Jordan matrices $U_{m,\alpha}$ (VII, p. 35, Prop. 8); then by Prop. 9 the integers m are all equal to 1 and so U is diagonal. Finally, the last assertion follows from the fact that if u is diagonalisable then its eigenspaces are the primary components of $E_u$, and from VII, p. 9, Cor. 1.