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Let $A$ be some linear operator(possibly over an infinite-dimensional space). Prove or deny that:

$$\ker\prod_{i = 1}^{k}(A - \lambda_i)^{d_i} = \sum_{i=1}^{k}\ker(A - \lambda_i)^{d_i}$$

Here $d_i, k $ are natural numbers, $\lambda_i $ are pairwise distinct real numbers.

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    $\begingroup$ It's true. It suffices to note that the operators $(A - \lambda_i)^{d_i}$ commute and have kernels that intersect trivially. $\endgroup$ – Omnomnomnom May 9 at 21:08

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