Cross product of two sets with vectors as elements.












1












$begingroup$


If $X in mathcal{X} = {x_1,x_2}$, for some $x_1,x_2 in mathbb{R}^{n times 1}$ and $Y in mathcal{Y} = {y_1,y_2}$, for some $y_1,y_2 in mathbb{R}^{m times 1}$, can I say that the stacked vector $Z = (X^T,Y^T)^T$ belongs to $mathcal{X} times mathcal{Y}$ ?



I think not, because $Z = (X^T,Y^T)^T in mathcal{Z} = left { begin{pmatrix} x_1 \ y_1 end{pmatrix}, begin{pmatrix} x_1 \ y_2 end{pmatrix}, begin{pmatrix} x_2 \ y_1 end{pmatrix}, begin{pmatrix} x_2 \ y_2 end{pmatrix} right}$, and



$
mathcal{X} times mathcal{Y} = { (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) }.
$



Then, is there a way to write $mathcal{Z}$ in terms if $mathcal{X}$ and $mathcal{Y}$ ?






linear-algebra discrete-mathematics







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    If $X in mathcal{X} = {x_1,x_2}$, for some $x_1,x_2 in mathbb{R}^{n times 1}$ and $Y in mathcal{Y} = {y_1,y_2}$, for some $y_1,y_2 in mathbb{R}^{m times 1}$, can I say that the stacked vector $Z = (X^T,Y^T)^T$ belongs to $mathcal{X} times mathcal{Y}$ ?



    I think not, because $Z = (X^T,Y^T)^T in mathcal{Z} = left { begin{pmatrix} x_1 \ y_1 end{pmatrix}, begin{pmatrix} x_1 \ y_2 end{pmatrix}, begin{pmatrix} x_2 \ y_1 end{pmatrix}, begin{pmatrix} x_2 \ y_2 end{pmatrix} right}$, and



    $
    mathcal{X} times mathcal{Y} = { (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) }.
    $



    Then, is there a way to write $mathcal{Z}$ in terms if $mathcal{X}$ and $mathcal{Y}$ ?






    linear-algebra discrete-mathematics







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      If $X in mathcal{X} = {x_1,x_2}$, for some $x_1,x_2 in mathbb{R}^{n times 1}$ and $Y in mathcal{Y} = {y_1,y_2}$, for some $y_1,y_2 in mathbb{R}^{m times 1}$, can I say that the stacked vector $Z = (X^T,Y^T)^T$ belongs to $mathcal{X} times mathcal{Y}$ ?



      I think not, because $Z = (X^T,Y^T)^T in mathcal{Z} = left { begin{pmatrix} x_1 \ y_1 end{pmatrix}, begin{pmatrix} x_1 \ y_2 end{pmatrix}, begin{pmatrix} x_2 \ y_1 end{pmatrix}, begin{pmatrix} x_2 \ y_2 end{pmatrix} right}$, and



      $
      mathcal{X} times mathcal{Y} = { (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) }.
      $



      Then, is there a way to write $mathcal{Z}$ in terms if $mathcal{X}$ and $mathcal{Y}$ ?






      linear-algebra discrete-mathematics







      share|cite|improve this question











      $endgroup$




      If $X in mathcal{X} = {x_1,x_2}$, for some $x_1,x_2 in mathbb{R}^{n times 1}$ and $Y in mathcal{Y} = {y_1,y_2}$, for some $y_1,y_2 in mathbb{R}^{m times 1}$, can I say that the stacked vector $Z = (X^T,Y^T)^T$ belongs to $mathcal{X} times mathcal{Y}$ ?



      I think not, because $Z = (X^T,Y^T)^T in mathcal{Z} = left { begin{pmatrix} x_1 \ y_1 end{pmatrix}, begin{pmatrix} x_1 \ y_2 end{pmatrix}, begin{pmatrix} x_2 \ y_1 end{pmatrix}, begin{pmatrix} x_2 \ y_2 end{pmatrix} right}$, and



      $
      mathcal{X} times mathcal{Y} = { (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) }.
      $



      Then, is there a way to write $mathcal{Z}$ in terms if $mathcal{X}$ and $mathcal{Y}$ ?






      linear-algebra discrete-mathematics




      linear-algebra discrete-mathematics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 25 at 5:15









      Henno Brandsma

      113k348122




      113k348122










      asked Jan 25 at 5:09









      OliVerOliVer

      718




      718






















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          $begingroup$

          Firstly, it seems that you've already defined the transpose $$(cdot)^T : bigcup_{n=1}^infty (mathbb{R}^{n}cupmathbb{R}^{n times 1}) rightarrow bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$$



          for all row and column vectors (I'm just using plain $mathbb{R}^n$ for the "rows" rather than $mathbb{R}^{1 times n}$).



          So the image of subsets $A subseteq bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$ under $(cdot)^T$ are defined already as



          $$(cdot)^T(A) = {x^T mid x in A}$$



          so for readability you might as well define



          $$A^T = (cdot)^T(A)$$



          This means that you can define $mathcal{Z}$ as




          $$mathcal{Z} = (mathcal{X}^T times mathcal{Y}^T)^T$$







          share|cite|improve this answer









          $endgroup$













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            0












            $begingroup$

            Firstly, it seems that you've already defined the transpose $$(cdot)^T : bigcup_{n=1}^infty (mathbb{R}^{n}cupmathbb{R}^{n times 1}) rightarrow bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$$



            for all row and column vectors (I'm just using plain $mathbb{R}^n$ for the "rows" rather than $mathbb{R}^{1 times n}$).



            So the image of subsets $A subseteq bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$ under $(cdot)^T$ are defined already as



            $$(cdot)^T(A) = {x^T mid x in A}$$



            so for readability you might as well define



            $$A^T = (cdot)^T(A)$$



            This means that you can define $mathcal{Z}$ as




            $$mathcal{Z} = (mathcal{X}^T times mathcal{Y}^T)^T$$







            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Firstly, it seems that you've already defined the transpose $$(cdot)^T : bigcup_{n=1}^infty (mathbb{R}^{n}cupmathbb{R}^{n times 1}) rightarrow bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$$



              for all row and column vectors (I'm just using plain $mathbb{R}^n$ for the "rows" rather than $mathbb{R}^{1 times n}$).



              So the image of subsets $A subseteq bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$ under $(cdot)^T$ are defined already as



              $$(cdot)^T(A) = {x^T mid x in A}$$



              so for readability you might as well define



              $$A^T = (cdot)^T(A)$$



              This means that you can define $mathcal{Z}$ as




              $$mathcal{Z} = (mathcal{X}^T times mathcal{Y}^T)^T$$







              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Firstly, it seems that you've already defined the transpose $$(cdot)^T : bigcup_{n=1}^infty (mathbb{R}^{n}cupmathbb{R}^{n times 1}) rightarrow bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$$



                for all row and column vectors (I'm just using plain $mathbb{R}^n$ for the "rows" rather than $mathbb{R}^{1 times n}$).



                So the image of subsets $A subseteq bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$ under $(cdot)^T$ are defined already as



                $$(cdot)^T(A) = {x^T mid x in A}$$



                so for readability you might as well define



                $$A^T = (cdot)^T(A)$$



                This means that you can define $mathcal{Z}$ as




                $$mathcal{Z} = (mathcal{X}^T times mathcal{Y}^T)^T$$







                share|cite|improve this answer









                $endgroup$



                Firstly, it seems that you've already defined the transpose $$(cdot)^T : bigcup_{n=1}^infty (mathbb{R}^{n}cupmathbb{R}^{n times 1}) rightarrow bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$$



                for all row and column vectors (I'm just using plain $mathbb{R}^n$ for the "rows" rather than $mathbb{R}^{1 times n}$).



                So the image of subsets $A subseteq bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$ under $(cdot)^T$ are defined already as



                $$(cdot)^T(A) = {x^T mid x in A}$$



                so for readability you might as well define



                $$A^T = (cdot)^T(A)$$



                This means that you can define $mathcal{Z}$ as




                $$mathcal{Z} = (mathcal{X}^T times mathcal{Y}^T)^T$$








                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 27 at 2:04









                MetricMetric

                1,25159




                1,25159






























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