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Concept of linear mappings are confusing me


Change of Basis ConfusionProve that a linear map for complex polynomials is diagonalizableEigenvalues of three given linear operatorsTransforming coordinate system vs objectsCan an $ntimes n$ matrix be reduced to a smaller matrix in any sense?Overview of Linear AlgebraLinear Transformation vs MatrixPruning SubsetsChange of basis formula - intuition/is this true?Linear Algebra:Vector Space













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!










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
    53 mins 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
    41 mins 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!










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
    53 mins 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
    41 mins 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!










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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 57 mins 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
    53 mins 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
    41 mins 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
    53 mins 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
    41 mins 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
53 mins 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
53 mins 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
41 mins 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
41 mins 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 $mathbbR^3$ and mapping it to a vector in $mathbbR^2$ 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
$$
beginbmatrix
2 & 0 & 1 \
3 & 4 & 0
endbmatrix
beginbmatrix
1 \
2 \
3
endbmatrix
=
beginbmatrix
5\
11
endbmatrix.
$$



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









    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      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 $mathbbR^3$ and mapping it to a vector in $mathbbR^2$ 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
      $$
      beginbmatrix
      2 & 0 & 1 \
      3 & 4 & 0
      endbmatrix
      beginbmatrix
      1 \
      2 \
      3
      endbmatrix
      =
      beginbmatrix
      5\
      11
      endbmatrix.
      $$



      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 $mathbbR^3$ and mapping it to a vector in $mathbbR^2$ 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
        $$
        beginbmatrix
        2 & 0 & 1 \
        3 & 4 & 0
        endbmatrix
        beginbmatrix
        1 \
        2 \
        3
        endbmatrix
        =
        beginbmatrix
        5\
        11
        endbmatrix.
        $$



        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 $mathbbR^3$ and mapping it to a vector in $mathbbR^2$ 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
          $$
          beginbmatrix
          2 & 0 & 1 \
          3 & 4 & 0
          endbmatrix
          beginbmatrix
          1 \
          2 \
          3
          endbmatrix
          =
          beginbmatrix
          5\
          11
          endbmatrix.
          $$



          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 $mathbbR^3$ and mapping it to a vector in $mathbbR^2$ 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
          $$
          beginbmatrix
          2 & 0 & 1 \
          3 & 4 & 0
          endbmatrix
          beginbmatrix
          1 \
          2 \
          3
          endbmatrix
          =
          beginbmatrix
          5\
          11
          endbmatrix.
          $$



          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 42 mins 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 32 mins ago









                  rschwiebrschwieb

                  108k12103253




                  108k12103253



























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