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












0














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




















  • 1




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










  • 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).
    – Mark Bennet
    Sep 28 '16 at 21:52










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
















0














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




















  • 1




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










  • 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).
    – Mark Bennet
    Sep 28 '16 at 21:52










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














0












0








0







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















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













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edited Sep 28 '16 at 22:28









Lonidard

2,9231921




2,9231921










asked Sep 28 '16 at 21:44









user5956

82




82








  • 1




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










  • 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).
    – Mark Bennet
    Sep 28 '16 at 21:52










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














  • 1




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










  • 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).
    – Mark Bennet
    Sep 28 '16 at 21:52










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








1




1




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




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












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).
– Mark Bennet
Sep 28 '16 at 21:52




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).
– Mark Bennet
Sep 28 '16 at 21:52












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




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










1 Answer
1






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oldest

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2














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






share|cite|improve this answer





















  • 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?
    – user5956
    Sep 28 '16 at 21:51












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










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











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






share|cite|improve this answer





















  • 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?
    – user5956
    Sep 28 '16 at 21:51












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










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
















2














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






share|cite|improve this answer





















  • 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?
    – user5956
    Sep 28 '16 at 21:51












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










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














2












2








2






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






share|cite|improve this answer












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 Pinilla

34.1k42771




34.1k42771












  • 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?
    – user5956
    Sep 28 '16 at 21:51












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










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


















  • 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?
    – user5956
    Sep 28 '16 at 21:51












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










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
















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?
– user5956
Sep 28 '16 at 21:51






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?
– user5956
Sep 28 '16 at 21:51














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




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












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




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


















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