Isomorphism and dimension exercise clarification.












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$begingroup$


I'd need a clarification on one of the statements here below.



Given that V is the set of linear operators f: $mathbb{R}^k rightarrow mathbb{R}^n$ , and I'm trying to prove that V has same dimension as $mathbb{R} ^{kn}$ so that to prove it.



Now, why is $ m=kn$ ? And also why is V contained in $(mathbb{R}^{n})^{(mathbb{R}^k)} ?$



I'm really looking for the reasoning (the simpler the better) behind these statements.



enter image description here










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$endgroup$












  • $begingroup$
    that argument is wrong! you need $f_i(A)subset B$, otherwise you are only modelling surjective morphisms.
    $endgroup$
    – Enkidu
    Dec 7 '18 at 12:03
















0












$begingroup$


I'd need a clarification on one of the statements here below.



Given that V is the set of linear operators f: $mathbb{R}^k rightarrow mathbb{R}^n$ , and I'm trying to prove that V has same dimension as $mathbb{R} ^{kn}$ so that to prove it.



Now, why is $ m=kn$ ? And also why is V contained in $(mathbb{R}^{n})^{(mathbb{R}^k)} ?$



I'm really looking for the reasoning (the simpler the better) behind these statements.



enter image description here










share|cite|improve this question









$endgroup$












  • $begingroup$
    that argument is wrong! you need $f_i(A)subset B$, otherwise you are only modelling surjective morphisms.
    $endgroup$
    – Enkidu
    Dec 7 '18 at 12:03














0












0








0





$begingroup$


I'd need a clarification on one of the statements here below.



Given that V is the set of linear operators f: $mathbb{R}^k rightarrow mathbb{R}^n$ , and I'm trying to prove that V has same dimension as $mathbb{R} ^{kn}$ so that to prove it.



Now, why is $ m=kn$ ? And also why is V contained in $(mathbb{R}^{n})^{(mathbb{R}^k)} ?$



I'm really looking for the reasoning (the simpler the better) behind these statements.



enter image description here










share|cite|improve this question









$endgroup$




I'd need a clarification on one of the statements here below.



Given that V is the set of linear operators f: $mathbb{R}^k rightarrow mathbb{R}^n$ , and I'm trying to prove that V has same dimension as $mathbb{R} ^{kn}$ so that to prove it.



Now, why is $ m=kn$ ? And also why is V contained in $(mathbb{R}^{n})^{(mathbb{R}^k)} ?$



I'm really looking for the reasoning (the simpler the better) behind these statements.



enter image description here







linear-algebra linear-transformations dimension-theory vector-space-isomorphism






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share|cite|improve this question











share|cite|improve this question




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asked Dec 7 '18 at 11:53









RicouelloRicouello

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1355












  • $begingroup$
    that argument is wrong! you need $f_i(A)subset B$, otherwise you are only modelling surjective morphisms.
    $endgroup$
    – Enkidu
    Dec 7 '18 at 12:03


















  • $begingroup$
    that argument is wrong! you need $f_i(A)subset B$, otherwise you are only modelling surjective morphisms.
    $endgroup$
    – Enkidu
    Dec 7 '18 at 12:03
















$begingroup$
that argument is wrong! you need $f_i(A)subset B$, otherwise you are only modelling surjective morphisms.
$endgroup$
– Enkidu
Dec 7 '18 at 12:03




$begingroup$
that argument is wrong! you need $f_i(A)subset B$, otherwise you are only modelling surjective morphisms.
$endgroup$
– Enkidu
Dec 7 '18 at 12:03










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