Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 1 / b / i / Solution

To prove the sum is direct, suppose ${\sum }_{j=1}^r{v}_j=0$ with $v_j\in V_{{\lambda }_j}$. Fix $i$ and apply the polynomial in the linear operatorEvery $v_j$ with $j\ne i$ is killed by one factor. On $v_i$, each factor is multiplication by ${\lambda }_i-{\lambda }_j$, givingThe scalar coefficient is nonzero because the eigenvalues are distinct. Thus $v_i=0$ for every $i$, provingIn particular, choosing a basis in each eigenspace yields a linearly independent combined list.