Mixing
Vector spaces and homogenous linear equations system
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For every vector space $R^n$ there is homogenous system of linear equations whose set of all solutions is isomorphic to $R^n$
This should obviously be true, but I am not sure I understand intuition behind it, for example how would I construct a system that is isomorphic, or rather generates $R^3$?
linear-algebra vector-spaces
asked Jan 19 at 23:27
DovlaDovla
869
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add a comment |
$begingroup$
For every vector space $R^n$ there is homogenous system of linear equations whose set of all solutions is isomorphic to $R^n$
This should obviously be true, but I am not sure I understand intuition behind it, for example how would I construct a system that is isomorphic, or rather generates $R^3$?
linear-algebra vector-spaces
asked Jan 19 at 23:27
DovlaDovla
869
$endgroup$
$begingroup$
If we represent the homogeneous system by the matrix equation $Ax=0$, then the only way the solution set is isomorphic to $mathbb R^n$ (as vector spaces) is if $A=0$.
$endgroup$
– Dave
Jan 19 at 23:31
$begingroup$
Its just identity. So, same identity generates all those spaces, but the difference in just the preset number of variables, or rather format of $x$? If that's the idea behind it.
$endgroup$
– Dovla
Jan 19 at 23:35
add a comment |
$begingroup$
For every vector space $R^n$ there is homogenous system of linear equations whose set of all solutions is isomorphic to $R^n$
This should obviously be true, but I am not sure I understand intuition behind it, for example how would I construct a system that is isomorphic, or rather generates $R^3$?
linear-algebra vector-spaces
asked Jan 19 at 23:27
DovlaDovla
869
$endgroup$
For every vector space $R^n$ there is homogenous system of linear equations whose set of all solutions is isomorphic to $R^n$
This should obviously be true, but I am not sure I understand intuition behind it, for example how would I construct a system that is isomorphic, or rather generates $R^3$?
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 19 at 23:27
DovlaDovla
869
asked Jan 19 at 23:27
DovlaDovla
869
asked Jan 19 at 23:27
DovlaDovla
869
asked Jan 19 at 23:27
DovlaDovla
869
asked Jan 19 at 23:27
DovlaDovla
869
869
$begingroup$
If we represent the homogeneous system by the matrix equation $Ax=0$, then the only way the solution set is isomorphic to $mathbb R^n$ (as vector spaces) is if $A=0$.
$endgroup$
– Dave
Jan 19 at 23:31
$begingroup$
Its just identity. So, same identity generates all those spaces, but the difference in just the preset number of variables, or rather format of $x$? If that's the idea behind it.
$endgroup$
– Dovla
Jan 19 at 23:35
add a comment |
$begingroup$
If we represent the homogeneous system by the matrix equation $Ax=0$, then the only way the solution set is isomorphic to $mathbb R^n$ (as vector spaces) is if $A=0$.
$endgroup$
– Dave
Jan 19 at 23:31
$begingroup$
Its just identity. So, same identity generates all those spaces, but the difference in just the preset number of variables, or rather format of $x$? If that's the idea behind it.
$endgroup$
– Dovla
Jan 19 at 23:35
$begingroup$
If we represent the homogeneous system by the matrix equation $Ax=0$, then the only way the solution set is isomorphic to $mathbb R^n$ (as vector spaces) is if $A=0$.
$endgroup$
– Dave
Jan 19 at 23:31
$begingroup$
If we represent the homogeneous system by the matrix equation $Ax=0$, then the only way the solution set is isomorphic to $mathbb R^n$ (as vector spaces) is if $A=0$.
$endgroup$
– Dave
Jan 19 at 23:31
$begingroup$
Its just identity. So, same identity generates all those spaces, but the difference in just the preset number of variables, or rather format of $x$? If that's the idea behind it.
$endgroup$
– Dovla
Jan 19 at 23:35
$begingroup$
Its just identity. So, same identity generates all those spaces, but the difference in just the preset number of variables, or rather format of $x$? If that's the idea behind it.
$endgroup$
– Dovla
Jan 19 at 23:35
add a comment |
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$begingroup$
If we represent the homogeneous system by the matrix equation $Ax=0$, then the only way the solution set is isomorphic to $mathbb R^n$ (as vector spaces) is if $A=0$.
$endgroup$
– Dave
Jan 19 at 23:31
$begingroup$
Its just identity. So, same identity generates all those spaces, but the difference in just the preset number of variables, or rather format of $x$? If that's the idea behind it.
$endgroup$
– Dovla
Jan 19 at 23:35