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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?

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Self-teaching worker
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    The eigenspace of an eigenvalue is completely unrelated to the algebraic multiplicity of that eigenvalue. If the eigenvalue is $\lambda$, the eigenspace is $ker(\lambda I - A)$ where $I$ denotes the $n\times n$ identity matrix. The algebraic multiplicity of $\lambda$ isn't used anywhere in this construction. – jflipp Nov 15 '14 at 18:39
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    That's what I missed: the eigenspace of $\lambda_i$ is the space $V_{\lambda_i}$ such that $AV_{\lambda_i}=\lambda_i V_{\lambda_i}$, but if the algebraic multiplicity of $\lambda_i$ is $>1$, the eigenspace is still one! Thank you so much!!! – Self-teaching worker Nov 15 '14 at 19:07

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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.