Question

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

Answer 1

I saw this question earlier and did the exact same proofs for scalar mult and addition closure

Answer 2

but what about showing that S is not empty ?

Answer 3

what about \(I\) that is a member, no?

Answer 4

I thought of the identity matrix as being the A vector, but is I the zero vector ?

Answer 5

doesn't make much sense for me :s

Answer 6

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?

Answer 7

yes

Answer 8

well the zero vector of a 4x4 matrix comes for free in R(nxn)
we already know it contains the zero vector

Answer 9

I don't know why I said 4x4, I meant nxn

Answer 10

\[\vec 0\in R_{n\times n}\]as a given