Concept of linear mappings are confusing me












5












$begingroup$


I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.



Thanks!






linear-algebra







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$endgroup$








  • 1




    $begingroup$
    maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
    $endgroup$
    – J. W. Tanner
    1 hour ago










  • $begingroup$
    Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
    $endgroup$
    – John Douma
    1 hour ago


















5












$begingroup$


I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.



Thanks!






linear-algebra







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
    $endgroup$
    – J. W. Tanner
    1 hour ago










  • $begingroup$
    Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
    $endgroup$
    – John Douma
    1 hour ago
















5












5








5





$begingroup$


I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.



Thanks!






linear-algebra







share|cite|improve this question









$endgroup$




I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.



Thanks!






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









mingming

4306




4306








  • 1




    $begingroup$
    maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
    $endgroup$
    – J. W. Tanner
    1 hour ago










  • $begingroup$
    Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
    $endgroup$
    – John Douma
    1 hour ago
















  • 1




    $begingroup$
    maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
    $endgroup$
    – J. W. Tanner
    1 hour ago










  • $begingroup$
    Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
    $endgroup$
    – John Douma
    1 hour ago










1




1




$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
1 hour ago




$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
1 hour ago












$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
1 hour ago






$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
1 hour ago












2 Answers
2






active

oldest

votes


















2












$begingroup$

For the moment don't think about multiplication and matrices.



You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$



Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
begin{bmatrix}
2 & 0 & 1 \
3 & 4 & 0
end{bmatrix}
begin{bmatrix}
1 \
2 \
3
end{bmatrix}
=
begin{bmatrix}
5\
11
end{bmatrix}.
$$



You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    A linear mapping has the property that it maps subspaces to subspaces.



    So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.



    By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

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      active

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      active

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      2












      $begingroup$

      For the moment don't think about multiplication and matrices.



      You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
      $$
      (x, y, z) mapsto (2x+ z, 3x+ 4y).
      $$



      Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
      $$
      begin{bmatrix}
      2 & 0 & 1 \
      3 & 4 & 0
      end{bmatrix}
      begin{bmatrix}
      1 \
      2 \
      3
      end{bmatrix}
      =
      begin{bmatrix}
      5\
      11
      end{bmatrix}.
      $$



      You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        For the moment don't think about multiplication and matrices.



        You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
        $$
        (x, y, z) mapsto (2x+ z, 3x+ 4y).
        $$



        Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
        $$
        begin{bmatrix}
        2 & 0 & 1 \
        3 & 4 & 0
        end{bmatrix}
        begin{bmatrix}
        1 \
        2 \
        3
        end{bmatrix}
        =
        begin{bmatrix}
        5\
        11
        end{bmatrix}.
        $$



        You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          For the moment don't think about multiplication and matrices.



          You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
          $$
          (x, y, z) mapsto (2x+ z, 3x+ 4y).
          $$



          Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
          $$
          begin{bmatrix}
          2 & 0 & 1 \
          3 & 4 & 0
          end{bmatrix}
          begin{bmatrix}
          1 \
          2 \
          3
          end{bmatrix}
          =
          begin{bmatrix}
          5\
          11
          end{bmatrix}.
          $$



          You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.






          share|cite|improve this answer









          $endgroup$



          For the moment don't think about multiplication and matrices.



          You can imagine starting from a vector $(x,y,z)$ in $mathbb{R}^3$ and mapping it to a vector in $mathbb{R}^2$ this way, for example:
          $$
          (x, y, z) mapsto (2x+ z, 3x+ 4y).
          $$



          Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
          $$
          begin{bmatrix}
          2 & 0 & 1 \
          3 & 4 & 0
          end{bmatrix}
          begin{bmatrix}
          1 \
          2 \
          3
          end{bmatrix}
          =
          begin{bmatrix}
          5\
          11
          end{bmatrix}.
          $$



          You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Ethan BolkerEthan Bolker

          45.8k553120




          45.8k553120























              0












              $begingroup$

              A linear mapping has the property that it maps subspaces to subspaces.



              So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.



              By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                A linear mapping has the property that it maps subspaces to subspaces.



                So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.



                By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  A linear mapping has the property that it maps subspaces to subspaces.



                  So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.



                  By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.






                  share|cite|improve this answer









                  $endgroup$



                  A linear mapping has the property that it maps subspaces to subspaces.



                  So it will map a line to a line or ${0}$, a plane to a plane, a line, or ${0}$, and so on.



                  By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  rschwiebrschwieb

                  108k12103253




                  108k12103253






























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