anonymous
  • anonymous
Let A be a fixed vector in R^(n*n) 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^(n*n) The work I did so far: if B in S then AB = BA => A (alpha B) = Alpha (AB) = Alpha (BA) => Alpha B is in S. Same closure under addition holds: if B and C are matrices in S then AB=BA and AC=CA => AB + AC = BA+CA => A(B+C) = (B+C)A => B + C in S. However, I don't know how to show that the set S is not empty, and i'm also not sure about what I did above
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TuringTest
  • TuringTest
I saw this question earlier and did the exact same proofs for scalar mult and addition closure
anonymous
  • anonymous
but what about showing that S is not empty ?
TuringTest
  • TuringTest
what about \(I\) that is a member, no?

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anonymous
  • anonymous
I thought of the identity matrix as being the A vector, but is I the zero vector ?
anonymous
  • anonymous
doesn't make much sense for me :s
TuringTest
  • TuringTest
People seem to have different standards about what to show for a subspace we all agree closure under addition and scalar multiplication are needed, but the other is just proving the existence of the zero vector, right?
anonymous
  • anonymous
yes
TuringTest
  • TuringTest
well the zero vector of a 4x4 matrix comes for free in R(nxn) we already know it contains the zero vector
TuringTest
  • TuringTest
I don't know why I said 4x4, I meant nxn
TuringTest
  • TuringTest
\[\vec 0\in R_{n\times n}\]as a given

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