I know, thanks to a kind user of this forum, that the sum of the eigenspaces of an endomorphism $A:V\to V$, with $\dim(V)=n$, is a direct sum.
A clear complete proof for the case where the eigenvalues of $A$ are distinct is here, for example.
I cannot manage to adapt that proof to the general case where $A$ may have eigenvalues of algebraic multiplicity $>1$. Could anybody explain a proof?
Best Answer
Let $\lambda_1,\ldots,\lambda_p$ the distinct eigenvalues of $A$ then since the polynomials $x-\lambda_k$ are $2$ by $2$ coprime then using the kernels decomposition theorem $^{(1)}$ we have that the sum of the eigenspaces of an endomorphism is direct.
$^{(1)}$ This is a french page and the result given there is a classic result but I didn't find the english translation of it and I'm not sure on what name this theorem is known in the English jargon.