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Proving M_mn(F) over F is a vector space

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Show M_mn(F) (the collection of mxn matrices over F) over a field F is a vector space.

    3. The attempt at a solution

    Denote [itex]A=a_{ij},B=b_{ij}[/itex] for elements of [itex]M_{mn}(F)[/itex] . Define [itex]A+B=(a_{ij}+b_{ij})[/itex] and for [itex]a\in F[/itex] denote [itex]\alpha A=(\alpha a_{ij})[/itex]. Then,

    (a) If [itex]\alpha\in F[/itex] and [itex]A\in M_{mn}(F)[/itex] ,[itex] \alpha A=(\alpha a_{ij})\in M_{mn}(F)[/itex] since [itex]F[/itex] is closed under scalar multiplication? not sure

    (b) If [itex]\alpha\in F and A,B\in M_{mn}(F)[/itex] , [itex]\alpha(A+B)=\alpha(a_{ij}+b_{ij})=\alpha a_{ij}+\alpha b_{ij}\in M_{mn}(F)[/itex] because of distributive property of [itex]F[/itex]?

    I'm kind of stuck in justifying the claims.
     
  2. jcsd
  3. Nov 27, 2011 #2
    Any ideas?
     
  4. Nov 28, 2011 #3

    Dick

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    The whole problem is more awkward to state than it actually is. Take the first one. alpha*a_ij isn't in F because F is closed under 'scalar multiplication'. It's because F is closed under multiplication in F. alpha is in F, a_ij is in F. So alpha*a_ij is in F. So alpha*A is in M_nm(F). There's nothing deep going on there.
     
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