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anonymous
 3 years ago
Let A be a fixed vector in R^(nxn) and let S be the set of all matrices that commute with A; that is,
S={B  AB=BA}
Show that S is a subspace of R^(nxn).
anonymous
 3 years ago
Let A be a fixed vector in R^(nxn) and let S be the set of all matrices that commute with A; that is, S={B  AB=BA} Show that S is a subspace of R^(nxn).

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what do you need to show in order to show something is a subspace of a vector space?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0All I know is that the subset has to satisfy two conditions. That is what I am not sure how to start.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think you only need to show two things 1) if \(w, v\in S\) then \(w+v\in S\) and 2) if \(w \in S, \lambda\in \mathbb{R}\) then \(\lambda w\in S\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i guess i should have written it with capital letters, but that is the idea

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you have two jobs 1) show that if \(B, C\) commute with \(A\), that is if \(AB=BA\) and \(AC=CA\) then \[A(B+C)=(B+C)A\] i.e. show that if \(B\in S\) and \(C\in S\) then \(B+C\in S\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this should be straight forward because of the distributive law

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you also have to show if \(AB=BA\) then \(A\lambda B=\lambda BA\) which again should be straight forward by the definition of scalar multiplication

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This answers every question I had. Thanks for taking your time. And then I just realized how simple this should've been.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0goal of course is to take the general definition and apply it in the specific case that is the hard part, rest is routine
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