Meaning behind $mathbf{V} subset mathbb{R}^n$












0












$begingroup$


When I have a vector space spanned by a set of vectors:$$mathbf{V}=mathrm{span}{(a,b,c,d)^T,(d,e,f,g)^T,(2a,2b,2c,2d)^T},$$How can I write $mathbf{V}$ in terms of a subset to another vector space? In this case $mathbf{V}$ is a subspace of dimension 2, but is it true that it is also a subspace of $mathbb{R}^3$ and $mathbb{R}^4$, i.e. $mathbf{V}subsetmathbb{R}^3$, $mathbf{V}subsetmathbb{R}^4$?



My confusion is that when I write $mathbf{V} subset mathbb{R}^3$, does it also imply the all of the element in $mathbf{V}$ are triples?






linear-algebra vector-spaces notation







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    When I have a vector space spanned by a set of vectors:$$mathbf{V}=mathrm{span}{(a,b,c,d)^T,(d,e,f,g)^T,(2a,2b,2c,2d)^T},$$How can I write $mathbf{V}$ in terms of a subset to another vector space? In this case $mathbf{V}$ is a subspace of dimension 2, but is it true that it is also a subspace of $mathbb{R}^3$ and $mathbb{R}^4$, i.e. $mathbf{V}subsetmathbb{R}^3$, $mathbf{V}subsetmathbb{R}^4$?



    My confusion is that when I write $mathbf{V} subset mathbb{R}^3$, does it also imply the all of the element in $mathbf{V}$ are triples?






    linear-algebra vector-spaces notation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      When I have a vector space spanned by a set of vectors:$$mathbf{V}=mathrm{span}{(a,b,c,d)^T,(d,e,f,g)^T,(2a,2b,2c,2d)^T},$$How can I write $mathbf{V}$ in terms of a subset to another vector space? In this case $mathbf{V}$ is a subspace of dimension 2, but is it true that it is also a subspace of $mathbb{R}^3$ and $mathbb{R}^4$, i.e. $mathbf{V}subsetmathbb{R}^3$, $mathbf{V}subsetmathbb{R}^4$?



      My confusion is that when I write $mathbf{V} subset mathbb{R}^3$, does it also imply the all of the element in $mathbf{V}$ are triples?






      linear-algebra vector-spaces notation







      share|cite|improve this question











      $endgroup$




      When I have a vector space spanned by a set of vectors:$$mathbf{V}=mathrm{span}{(a,b,c,d)^T,(d,e,f,g)^T,(2a,2b,2c,2d)^T},$$How can I write $mathbf{V}$ in terms of a subset to another vector space? In this case $mathbf{V}$ is a subspace of dimension 2, but is it true that it is also a subspace of $mathbb{R}^3$ and $mathbb{R}^4$, i.e. $mathbf{V}subsetmathbb{R}^3$, $mathbf{V}subsetmathbb{R}^4$?



      My confusion is that when I write $mathbf{V} subset mathbb{R}^3$, does it also imply the all of the element in $mathbf{V}$ are triples?






      linear-algebra vector-spaces notation




      linear-algebra vector-spaces notation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 15 '18 at 3:52









      Saad

      19.7k92352




      19.7k92352










      asked Dec 15 '18 at 3:43









      WeiShan NgWeiShan Ng

      617




      617






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Yes. $Vsubsetmathbb R^4$. But $Vnotsubsetmathbb R^3$, here.



          Though $V$ is $2$-dimensional, it is a subspace of $mathbb R^4$. Vectors in $mathbb R^4$ are expressed as $4$-tuples.



          Whereas $mathbb R^3$ "sits" in $mathbb R^4$ in a fairly obvious way, we cannot say, for instance, that any $2$-dimensional subspace of $mathbb R^4$ is a subspace of $mathbb R^3$. If the fourth coordinate is $0$ for all elements of $V$, we can say that, in a sense, $Vsubset mathbb R^3$. But here we would be thinking of $mathbb R^3$ as the set of $4$-tuples with fourth coordinate $0$. Note there are other $"mathbb R^3"$s in $mathbb R^4$ (let each of the other $3$ coordinates be $0$).



          Plus there are lots of other there $3$-dimensional subspaces (one corresponding to each normal direction).



          Btw, all $n$-dimensional vector spaces are isomorphic.



          All this can take some getting used to.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So we can't simply use the notation $mathbf{V} subseteq mathbb{R}^n$ to say that the subspace $mathbf{V}$ has n dimension, because its element might be of $(geqslant n)$-tuples?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:40












          • $begingroup$
            Right. Any higher dimensional space has $n$-dimensional subspaces.
            $endgroup$
            – Chris Custer
            Dec 15 '18 at 4:42



















          0












          $begingroup$

          No. For example , span of (1,1) in a plane is the line passes through the origin which makes an angle of 45 degree in each of the coordinate axes. It is one dimensional. But considering the same set in a three dimensional space is meaningless.
          Instead we consider span(1,1,0).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does it mean I can only say $mathrm{span} { (1,1)^T } subset mathbb{R}^2$, but not $mathrm{span} { (1,1)^T } subset mathbb{R}^3$?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:08












          • $begingroup$
            Yes. The second one is meaningless
            $endgroup$
            – Chinnapparaj R
            Dec 15 '18 at 4:10










          • $begingroup$
            I'm very confused with the notation, how do we represent a vector space $mathbf{V}$ that is of dimension $n$, but having elements of m-tuples ? Like how do we represent $mathbf{V}=mathrm{span} {(1,1,0,0)^T,(1,0,1,0)^T,(1,0,0,1)^T }$? Do I just simply write it as $mathbf{V} in mathbb{R}^3$?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:13












          • $begingroup$
            The vectors are in a four dimensional space so you can say V is a subset of a four dimensional space
            $endgroup$
            – Chinnapparaj R
            Dec 15 '18 at 4:22










          • $begingroup$
            I think I get it now, thank you!
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:27











          Your Answer





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          2 Answers
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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Yes. $Vsubsetmathbb R^4$. But $Vnotsubsetmathbb R^3$, here.



          Though $V$ is $2$-dimensional, it is a subspace of $mathbb R^4$. Vectors in $mathbb R^4$ are expressed as $4$-tuples.



          Whereas $mathbb R^3$ "sits" in $mathbb R^4$ in a fairly obvious way, we cannot say, for instance, that any $2$-dimensional subspace of $mathbb R^4$ is a subspace of $mathbb R^3$. If the fourth coordinate is $0$ for all elements of $V$, we can say that, in a sense, $Vsubset mathbb R^3$. But here we would be thinking of $mathbb R^3$ as the set of $4$-tuples with fourth coordinate $0$. Note there are other $"mathbb R^3"$s in $mathbb R^4$ (let each of the other $3$ coordinates be $0$).



          Plus there are lots of other there $3$-dimensional subspaces (one corresponding to each normal direction).



          Btw, all $n$-dimensional vector spaces are isomorphic.



          All this can take some getting used to.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So we can't simply use the notation $mathbf{V} subseteq mathbb{R}^n$ to say that the subspace $mathbf{V}$ has n dimension, because its element might be of $(geqslant n)$-tuples?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:40












          • $begingroup$
            Right. Any higher dimensional space has $n$-dimensional subspaces.
            $endgroup$
            – Chris Custer
            Dec 15 '18 at 4:42
















          1












          $begingroup$

          Yes. $Vsubsetmathbb R^4$. But $Vnotsubsetmathbb R^3$, here.



          Though $V$ is $2$-dimensional, it is a subspace of $mathbb R^4$. Vectors in $mathbb R^4$ are expressed as $4$-tuples.



          Whereas $mathbb R^3$ "sits" in $mathbb R^4$ in a fairly obvious way, we cannot say, for instance, that any $2$-dimensional subspace of $mathbb R^4$ is a subspace of $mathbb R^3$. If the fourth coordinate is $0$ for all elements of $V$, we can say that, in a sense, $Vsubset mathbb R^3$. But here we would be thinking of $mathbb R^3$ as the set of $4$-tuples with fourth coordinate $0$. Note there are other $"mathbb R^3"$s in $mathbb R^4$ (let each of the other $3$ coordinates be $0$).



          Plus there are lots of other there $3$-dimensional subspaces (one corresponding to each normal direction).



          Btw, all $n$-dimensional vector spaces are isomorphic.



          All this can take some getting used to.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So we can't simply use the notation $mathbf{V} subseteq mathbb{R}^n$ to say that the subspace $mathbf{V}$ has n dimension, because its element might be of $(geqslant n)$-tuples?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:40












          • $begingroup$
            Right. Any higher dimensional space has $n$-dimensional subspaces.
            $endgroup$
            – Chris Custer
            Dec 15 '18 at 4:42














          1












          1








          1





          $begingroup$

          Yes. $Vsubsetmathbb R^4$. But $Vnotsubsetmathbb R^3$, here.



          Though $V$ is $2$-dimensional, it is a subspace of $mathbb R^4$. Vectors in $mathbb R^4$ are expressed as $4$-tuples.



          Whereas $mathbb R^3$ "sits" in $mathbb R^4$ in a fairly obvious way, we cannot say, for instance, that any $2$-dimensional subspace of $mathbb R^4$ is a subspace of $mathbb R^3$. If the fourth coordinate is $0$ for all elements of $V$, we can say that, in a sense, $Vsubset mathbb R^3$. But here we would be thinking of $mathbb R^3$ as the set of $4$-tuples with fourth coordinate $0$. Note there are other $"mathbb R^3"$s in $mathbb R^4$ (let each of the other $3$ coordinates be $0$).



          Plus there are lots of other there $3$-dimensional subspaces (one corresponding to each normal direction).



          Btw, all $n$-dimensional vector spaces are isomorphic.



          All this can take some getting used to.






          share|cite|improve this answer









          $endgroup$



          Yes. $Vsubsetmathbb R^4$. But $Vnotsubsetmathbb R^3$, here.



          Though $V$ is $2$-dimensional, it is a subspace of $mathbb R^4$. Vectors in $mathbb R^4$ are expressed as $4$-tuples.



          Whereas $mathbb R^3$ "sits" in $mathbb R^4$ in a fairly obvious way, we cannot say, for instance, that any $2$-dimensional subspace of $mathbb R^4$ is a subspace of $mathbb R^3$. If the fourth coordinate is $0$ for all elements of $V$, we can say that, in a sense, $Vsubset mathbb R^3$. But here we would be thinking of $mathbb R^3$ as the set of $4$-tuples with fourth coordinate $0$. Note there are other $"mathbb R^3"$s in $mathbb R^4$ (let each of the other $3$ coordinates be $0$).



          Plus there are lots of other there $3$-dimensional subspaces (one corresponding to each normal direction).



          Btw, all $n$-dimensional vector spaces are isomorphic.



          All this can take some getting used to.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 4:17









          Chris CusterChris Custer

          13.6k3827




          13.6k3827












          • $begingroup$
            So we can't simply use the notation $mathbf{V} subseteq mathbb{R}^n$ to say that the subspace $mathbf{V}$ has n dimension, because its element might be of $(geqslant n)$-tuples?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:40












          • $begingroup$
            Right. Any higher dimensional space has $n$-dimensional subspaces.
            $endgroup$
            – Chris Custer
            Dec 15 '18 at 4:42


















          • $begingroup$
            So we can't simply use the notation $mathbf{V} subseteq mathbb{R}^n$ to say that the subspace $mathbf{V}$ has n dimension, because its element might be of $(geqslant n)$-tuples?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:40












          • $begingroup$
            Right. Any higher dimensional space has $n$-dimensional subspaces.
            $endgroup$
            – Chris Custer
            Dec 15 '18 at 4:42
















          $begingroup$
          So we can't simply use the notation $mathbf{V} subseteq mathbb{R}^n$ to say that the subspace $mathbf{V}$ has n dimension, because its element might be of $(geqslant n)$-tuples?
          $endgroup$
          – WeiShan Ng
          Dec 15 '18 at 4:40






          $begingroup$
          So we can't simply use the notation $mathbf{V} subseteq mathbb{R}^n$ to say that the subspace $mathbf{V}$ has n dimension, because its element might be of $(geqslant n)$-tuples?
          $endgroup$
          – WeiShan Ng
          Dec 15 '18 at 4:40














          $begingroup$
          Right. Any higher dimensional space has $n$-dimensional subspaces.
          $endgroup$
          – Chris Custer
          Dec 15 '18 at 4:42




          $begingroup$
          Right. Any higher dimensional space has $n$-dimensional subspaces.
          $endgroup$
          – Chris Custer
          Dec 15 '18 at 4:42











          0












          $begingroup$

          No. For example , span of (1,1) in a plane is the line passes through the origin which makes an angle of 45 degree in each of the coordinate axes. It is one dimensional. But considering the same set in a three dimensional space is meaningless.
          Instead we consider span(1,1,0).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does it mean I can only say $mathrm{span} { (1,1)^T } subset mathbb{R}^2$, but not $mathrm{span} { (1,1)^T } subset mathbb{R}^3$?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:08












          • $begingroup$
            Yes. The second one is meaningless
            $endgroup$
            – Chinnapparaj R
            Dec 15 '18 at 4:10










          • $begingroup$
            I'm very confused with the notation, how do we represent a vector space $mathbf{V}$ that is of dimension $n$, but having elements of m-tuples ? Like how do we represent $mathbf{V}=mathrm{span} {(1,1,0,0)^T,(1,0,1,0)^T,(1,0,0,1)^T }$? Do I just simply write it as $mathbf{V} in mathbb{R}^3$?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:13












          • $begingroup$
            The vectors are in a four dimensional space so you can say V is a subset of a four dimensional space
            $endgroup$
            – Chinnapparaj R
            Dec 15 '18 at 4:22










          • $begingroup$
            I think I get it now, thank you!
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:27
















          0












          $begingroup$

          No. For example , span of (1,1) in a plane is the line passes through the origin which makes an angle of 45 degree in each of the coordinate axes. It is one dimensional. But considering the same set in a three dimensional space is meaningless.
          Instead we consider span(1,1,0).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does it mean I can only say $mathrm{span} { (1,1)^T } subset mathbb{R}^2$, but not $mathrm{span} { (1,1)^T } subset mathbb{R}^3$?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:08












          • $begingroup$
            Yes. The second one is meaningless
            $endgroup$
            – Chinnapparaj R
            Dec 15 '18 at 4:10










          • $begingroup$
            I'm very confused with the notation, how do we represent a vector space $mathbf{V}$ that is of dimension $n$, but having elements of m-tuples ? Like how do we represent $mathbf{V}=mathrm{span} {(1,1,0,0)^T,(1,0,1,0)^T,(1,0,0,1)^T }$? Do I just simply write it as $mathbf{V} in mathbb{R}^3$?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:13












          • $begingroup$
            The vectors are in a four dimensional space so you can say V is a subset of a four dimensional space
            $endgroup$
            – Chinnapparaj R
            Dec 15 '18 at 4:22










          • $begingroup$
            I think I get it now, thank you!
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:27














          0












          0








          0





          $begingroup$

          No. For example , span of (1,1) in a plane is the line passes through the origin which makes an angle of 45 degree in each of the coordinate axes. It is one dimensional. But considering the same set in a three dimensional space is meaningless.
          Instead we consider span(1,1,0).






          share|cite|improve this answer









          $endgroup$



          No. For example , span of (1,1) in a plane is the line passes through the origin which makes an angle of 45 degree in each of the coordinate axes. It is one dimensional. But considering the same set in a three dimensional space is meaningless.
          Instead we consider span(1,1,0).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 4:02









          Chinnapparaj RChinnapparaj R

          5,5072928




          5,5072928












          • $begingroup$
            Does it mean I can only say $mathrm{span} { (1,1)^T } subset mathbb{R}^2$, but not $mathrm{span} { (1,1)^T } subset mathbb{R}^3$?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:08












          • $begingroup$
            Yes. The second one is meaningless
            $endgroup$
            – Chinnapparaj R
            Dec 15 '18 at 4:10










          • $begingroup$
            I'm very confused with the notation, how do we represent a vector space $mathbf{V}$ that is of dimension $n$, but having elements of m-tuples ? Like how do we represent $mathbf{V}=mathrm{span} {(1,1,0,0)^T,(1,0,1,0)^T,(1,0,0,1)^T }$? Do I just simply write it as $mathbf{V} in mathbb{R}^3$?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:13












          • $begingroup$
            The vectors are in a four dimensional space so you can say V is a subset of a four dimensional space
            $endgroup$
            – Chinnapparaj R
            Dec 15 '18 at 4:22










          • $begingroup$
            I think I get it now, thank you!
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:27


















          • $begingroup$
            Does it mean I can only say $mathrm{span} { (1,1)^T } subset mathbb{R}^2$, but not $mathrm{span} { (1,1)^T } subset mathbb{R}^3$?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:08












          • $begingroup$
            Yes. The second one is meaningless
            $endgroup$
            – Chinnapparaj R
            Dec 15 '18 at 4:10










          • $begingroup$
            I'm very confused with the notation, how do we represent a vector space $mathbf{V}$ that is of dimension $n$, but having elements of m-tuples ? Like how do we represent $mathbf{V}=mathrm{span} {(1,1,0,0)^T,(1,0,1,0)^T,(1,0,0,1)^T }$? Do I just simply write it as $mathbf{V} in mathbb{R}^3$?
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:13












          • $begingroup$
            The vectors are in a four dimensional space so you can say V is a subset of a four dimensional space
            $endgroup$
            – Chinnapparaj R
            Dec 15 '18 at 4:22










          • $begingroup$
            I think I get it now, thank you!
            $endgroup$
            – WeiShan Ng
            Dec 15 '18 at 4:27
















          $begingroup$
          Does it mean I can only say $mathrm{span} { (1,1)^T } subset mathbb{R}^2$, but not $mathrm{span} { (1,1)^T } subset mathbb{R}^3$?
          $endgroup$
          – WeiShan Ng
          Dec 15 '18 at 4:08






          $begingroup$
          Does it mean I can only say $mathrm{span} { (1,1)^T } subset mathbb{R}^2$, but not $mathrm{span} { (1,1)^T } subset mathbb{R}^3$?
          $endgroup$
          – WeiShan Ng
          Dec 15 '18 at 4:08














          $begingroup$
          Yes. The second one is meaningless
          $endgroup$
          – Chinnapparaj R
          Dec 15 '18 at 4:10




          $begingroup$
          Yes. The second one is meaningless
          $endgroup$
          – Chinnapparaj R
          Dec 15 '18 at 4:10












          $begingroup$
          I'm very confused with the notation, how do we represent a vector space $mathbf{V}$ that is of dimension $n$, but having elements of m-tuples ? Like how do we represent $mathbf{V}=mathrm{span} {(1,1,0,0)^T,(1,0,1,0)^T,(1,0,0,1)^T }$? Do I just simply write it as $mathbf{V} in mathbb{R}^3$?
          $endgroup$
          – WeiShan Ng
          Dec 15 '18 at 4:13






          $begingroup$
          I'm very confused with the notation, how do we represent a vector space $mathbf{V}$ that is of dimension $n$, but having elements of m-tuples ? Like how do we represent $mathbf{V}=mathrm{span} {(1,1,0,0)^T,(1,0,1,0)^T,(1,0,0,1)^T }$? Do I just simply write it as $mathbf{V} in mathbb{R}^3$?
          $endgroup$
          – WeiShan Ng
          Dec 15 '18 at 4:13














          $begingroup$
          The vectors are in a four dimensional space so you can say V is a subset of a four dimensional space
          $endgroup$
          – Chinnapparaj R
          Dec 15 '18 at 4:22




          $begingroup$
          The vectors are in a four dimensional space so you can say V is a subset of a four dimensional space
          $endgroup$
          – Chinnapparaj R
          Dec 15 '18 at 4:22












          $begingroup$
          I think I get it now, thank you!
          $endgroup$
          – WeiShan Ng
          Dec 15 '18 at 4:27




          $begingroup$
          I think I get it now, thank you!
          $endgroup$
          – WeiShan Ng
          Dec 15 '18 at 4:27


















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          How do I know what Microsoft account the skydrive app is syncing to?

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          up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
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          Grease: Live!

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          A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
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          When does type information flow backwards in C++?

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          up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
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