Is null(T-\lambdaI) invariant under S for every \lambda \in F?

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To prove that null(T - λI) is invariant under S for every λ in F, it is essential to demonstrate that if x is in null(T - λI), then Sx also belongs to null(T - λI). This means showing that if Tx = λx, then S(Tx) = S(λx). Since S and T commute (ST = TS), we can express S(Tx) as TSx, leading to TSx = λSx. Thus, if x satisfies the property, Sx will also satisfy it, confirming the invariance of null(T - λI) under S. The proof hinges on the commutativity of S and T.
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Suppose S, T \in L(V) are such that ST=TS. Prove that null(T-\lambdaI) is invariant under S for every \lambda \in F.


I can't find anything helpful in my book, I'm not sure where to start...
 
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null(T-lambda*I) is just the set of all vectors x such that (T-lambda*I)x=0. So Tx-lambda*x=0. Or Tx=lambda*x. To show this is invariant under S you just have to show that if x satisfies that property, then so does Sx.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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