If a linear operator T makes a vector zero, a product of linear operators containing T makes that vector zero. Why?

Question

I am reading a linear algebra book and I don't understand some part of it. The chapter is about diagonalization of linear operators.I state what is exactly in the book:

Let $T$ be a diagonizable linear operator and let $c_1,...,c_k$ be the distinct characteristic values of $T$. Then it is easy to see that the minimal polynomial fot $T$ is the polynomial $$p=(x-c_1)...(x-c_k).$$ If $\alpha$ is a characteristic vector, then one of the operators $T-c_1I,...,T-c_kI$ sends $\alpha$ into $0$. Therefore $$(T-c_1I)...(T-c_kI)\alpha=0$$ for every characteristic vector $\alpha$.

Please explain why the last claim is true i.e. $(T-c_1I)...(T-c_kI)\alpha=0$ for every characteristic vector $\alpha$.

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I got my answer. My answer is the duplicate of this and the answer is the commutativity of the operators (T-c_kI).

Answer 1

HINT. The operators $T-c_1I, T-c_2I, \ldots , T-c_k I$ commute. This means that you can change the order in the following product: $$ (T-c_1I)(T-c_2I) \ldots (T-c_k I).$$