quiz3-solns - T-λI call it v So Sv = av for some scalar a...

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MATH 110 - SOLUTION TO QUIZ 3 LECTURE 1, SUMMER 2009 GSI: SANTIAGO CA ˜ NEZ Suppose that V is a vector space over a field F and that S,T ∈ L ( V ) are operators such that ST = TS . Prove that null( T - λI ) is S -invariant for any λ F . Proof. Let λ F and let v null( T - λI ). Then ( T - λI ) v = Tv - λv = 0 . We claim that Sv null( T - λI ). Indeed, keeping in mind that ST = TS , we have ( T - λI )( Sv ) = TSv - λSv = STv - S ( λv ) = S ( Tv - λv ) = S (0) = 0 . Thus Sv null( T - λI ), showing that null( T - λI ) is S -invariant. ± Just a quick remark about what this result really means. Suppose that F is actually C and that V is finite-dimensional. If λ is an eigenvalue of T , this result says that S becomes an operator on the eigenspace null( T - λI ). Thus S , being an operator on a finite-dimensional complex vector space, has an eigenvector in null(
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Unformatted text preview: T-λI ), call it v . So Sv = av for some scalar a ∈ C , which is not necessarily λ itself. But since v comes from null( T-λI ), we also have Tv = λv, so v is also an eigenvector of T . We conclude that two commuting operators have a common eigenvector. Of course, the eigenvalues which correspond to this common eigenvector are not necessarily the same for S and T , but that’s okay — the important thing is the existence of a common eigenvector. We will see later that this fact, in some ways, explains the way the universe works :) Date : July 9, 2009....
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