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Proving M_mn(F) over F is a vector space

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Homework Statement



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

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.
 

Answers and Replies

  • #2
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Any ideas?
 
  • #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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