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

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          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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          Probability when a professor distributes a quiz and homework assignment to a class of n students.

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          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
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          Aardman Animations

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          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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          Are they similar matrix

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          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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