Isomorphism and dimension exercise clarification.
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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.

linear-algebra linear-transformations dimension-theory vector-space-isomorphism
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add a comment |
$begingroup$
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.

linear-algebra linear-transformations dimension-theory vector-space-isomorphism
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.

linear-algebra linear-transformations dimension-theory vector-space-isomorphism
$endgroup$
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.

linear-algebra linear-transformations dimension-theory vector-space-isomorphism
linear-algebra linear-transformations dimension-theory vector-space-isomorphism
asked Dec 7 '18 at 11:53
RicouelloRicouello
1355
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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