# Prove or deny that $\ker\prod_{i = 1}^{k}(A - \lambda_i)^{d_i} = \sum_{i=1}^{k}\ker(A - \lambda_i)^{d_i}.$

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.

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