Suppose $V, $and $W$ are both finite dimensional . Prove that there exits an injective linear map from $V$ to...












0












$begingroup$


Suppose $V$ and $W$ are both finite dimensional. Prove that there exits an injective linear map from $V$ to $W$ if and only if $dim V leq dim W$.



Proof: Suppose $V$ and $W$ are both finite dimensional. Then we have $dim V = dim(text{null} (T)) + dim (text{range} (T))$. Then we have $dim V= dim text{range} (T)leq dim W$.
So there exists an injective map, so that $dim V leq dim W$.



Conversely, suppose $dim Vleq dim W$. Then let $dim V = n, dim W = m$. Define a basis for $V$ by $v_1,ldots,v_n $ for $V$ and $w_1,ldots,w_m$ for $W$.
Can someone please help me? I dont really know how to show there is an injective linear map .






linear-algebra







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$endgroup$








  • 1




    $begingroup$
    You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
    $endgroup$
    – anderstood
    Sep 28 '16 at 21:47










  • $begingroup$
    Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
    $endgroup$
    – Mark Bennet
    Sep 28 '16 at 21:52










  • $begingroup$
    $T$ can be defined as $T: V→W$
    $endgroup$
    – user5956
    Sep 28 '16 at 21:53
















0












$begingroup$


Suppose $V$ and $W$ are both finite dimensional. Prove that there exits an injective linear map from $V$ to $W$ if and only if $dim V leq dim W$.



Proof: Suppose $V$ and $W$ are both finite dimensional. Then we have $dim V = dim(text{null} (T)) + dim (text{range} (T))$. Then we have $dim V= dim text{range} (T)leq dim W$.
So there exists an injective map, so that $dim V leq dim W$.



Conversely, suppose $dim Vleq dim W$. Then let $dim V = n, dim W = m$. Define a basis for $V$ by $v_1,ldots,v_n $ for $V$ and $w_1,ldots,w_m$ for $W$.
Can someone please help me? I dont really know how to show there is an injective linear map .






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
    $endgroup$
    – anderstood
    Sep 28 '16 at 21:47










  • $begingroup$
    Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
    $endgroup$
    – Mark Bennet
    Sep 28 '16 at 21:52










  • $begingroup$
    $T$ can be defined as $T: V→W$
    $endgroup$
    – user5956
    Sep 28 '16 at 21:53














0












0








0





$begingroup$


Suppose $V$ and $W$ are both finite dimensional. Prove that there exits an injective linear map from $V$ to $W$ if and only if $dim V leq dim W$.



Proof: Suppose $V$ and $W$ are both finite dimensional. Then we have $dim V = dim(text{null} (T)) + dim (text{range} (T))$. Then we have $dim V= dim text{range} (T)leq dim W$.
So there exists an injective map, so that $dim V leq dim W$.



Conversely, suppose $dim Vleq dim W$. Then let $dim V = n, dim W = m$. Define a basis for $V$ by $v_1,ldots,v_n $ for $V$ and $w_1,ldots,w_m$ for $W$.
Can someone please help me? I dont really know how to show there is an injective linear map .






linear-algebra







share|cite|improve this question











$endgroup$




Suppose $V$ and $W$ are both finite dimensional. Prove that there exits an injective linear map from $V$ to $W$ if and only if $dim V leq dim W$.



Proof: Suppose $V$ and $W$ are both finite dimensional. Then we have $dim V = dim(text{null} (T)) + dim (text{range} (T))$. Then we have $dim V= dim text{range} (T)leq dim W$.
So there exists an injective map, so that $dim V leq dim W$.



Conversely, suppose $dim Vleq dim W$. Then let $dim V = n, dim W = m$. Define a basis for $V$ by $v_1,ldots,v_n $ for $V$ and $w_1,ldots,w_m$ for $W$.
Can someone please help me? I dont really know how to show there is an injective linear map .






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 28 '16 at 22:28









Lonidard

2,93311021




2,93311021










asked Sep 28 '16 at 21:44









user5956user5956

82




82








  • 1




    $begingroup$
    You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
    $endgroup$
    – anderstood
    Sep 28 '16 at 21:47










  • $begingroup$
    Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
    $endgroup$
    – Mark Bennet
    Sep 28 '16 at 21:52










  • $begingroup$
    $T$ can be defined as $T: V→W$
    $endgroup$
    – user5956
    Sep 28 '16 at 21:53














  • 1




    $begingroup$
    You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
    $endgroup$
    – anderstood
    Sep 28 '16 at 21:47










  • $begingroup$
    Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
    $endgroup$
    – Mark Bennet
    Sep 28 '16 at 21:52










  • $begingroup$
    $T$ can be defined as $T: V→W$
    $endgroup$
    – user5956
    Sep 28 '16 at 21:53








1




1




$begingroup$
You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
$endgroup$
– anderstood
Sep 28 '16 at 21:47




$begingroup$
You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
$endgroup$
– anderstood
Sep 28 '16 at 21:47












$begingroup$
Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
$endgroup$
– Mark Bennet
Sep 28 '16 at 21:52




$begingroup$
Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
$endgroup$
– Mark Bennet
Sep 28 '16 at 21:52












$begingroup$
$T$ can be defined as $T: V→W$
$endgroup$
– user5956
Sep 28 '16 at 21:53




$begingroup$
$T$ can be defined as $T: V→W$
$endgroup$
– user5956
Sep 28 '16 at 21:53










1 Answer
1






active

oldest

votes


















2












$begingroup$

Hint: define $T$ by $v_ilongmapsto w_i$, $i=1,dots n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
    $endgroup$
    – user5956
    Sep 28 '16 at 21:51












  • $begingroup$
    HI Martin , can you please give me some feedback? thanks
    $endgroup$
    – user5956
    Sep 28 '16 at 23:27










  • $begingroup$
    @user5956, right.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Sep 29 '16 at 7:40











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Hint: define $T$ by $v_ilongmapsto w_i$, $i=1,dots n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
    $endgroup$
    – user5956
    Sep 28 '16 at 21:51












  • $begingroup$
    HI Martin , can you please give me some feedback? thanks
    $endgroup$
    – user5956
    Sep 28 '16 at 23:27










  • $begingroup$
    @user5956, right.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Sep 29 '16 at 7:40
















2












$begingroup$

Hint: define $T$ by $v_ilongmapsto w_i$, $i=1,dots n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
    $endgroup$
    – user5956
    Sep 28 '16 at 21:51












  • $begingroup$
    HI Martin , can you please give me some feedback? thanks
    $endgroup$
    – user5956
    Sep 28 '16 at 23:27










  • $begingroup$
    @user5956, right.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Sep 29 '16 at 7:40














2












2








2





$begingroup$

Hint: define $T$ by $v_ilongmapsto w_i$, $i=1,dots n$.






share|cite|improve this answer









$endgroup$



Hint: define $T$ by $v_ilongmapsto w_i$, $i=1,dots n$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 28 '16 at 21:47









Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.2k42871




34.2k42871












  • $begingroup$
    So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
    $endgroup$
    – user5956
    Sep 28 '16 at 21:51












  • $begingroup$
    HI Martin , can you please give me some feedback? thanks
    $endgroup$
    – user5956
    Sep 28 '16 at 23:27










  • $begingroup$
    @user5956, right.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Sep 29 '16 at 7:40


















  • $begingroup$
    So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
    $endgroup$
    – user5956
    Sep 28 '16 at 21:51












  • $begingroup$
    HI Martin , can you please give me some feedback? thanks
    $endgroup$
    – user5956
    Sep 28 '16 at 23:27










  • $begingroup$
    @user5956, right.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Sep 29 '16 at 7:40
















$begingroup$
So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
$endgroup$
– user5956
Sep 28 '16 at 21:51






$begingroup$
So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
$endgroup$
– user5956
Sep 28 '16 at 21:51














$begingroup$
HI Martin , can you please give me some feedback? thanks
$endgroup$
– user5956
Sep 28 '16 at 23:27




$begingroup$
HI Martin , can you please give me some feedback? thanks
$endgroup$
– user5956
Sep 28 '16 at 23:27












$begingroup$
@user5956, right.
$endgroup$
– Martín-Blas Pérez Pinilla
Sep 29 '16 at 7:40




$begingroup$
@user5956, right.
$endgroup$
– Martín-Blas Pérez Pinilla
Sep 29 '16 at 7:40


















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