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

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The discussion centers on proving that the null space of the operator \( T - \lambda I \) is invariant under another operator \( S \) when \( S \) and \( T \) commute, i.e., \( ST = TS \). The null space is defined as the set of all vectors \( x \) such that \( (T - \lambda I)x = 0 \), which simplifies to \( Tx = \lambda x \). The proof requires demonstrating that if \( x \) is in the null space, then \( Sx \) also satisfies the same property, confirming the invariance.

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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.
 

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