Geometric meaning of left multiply with a diagonal matrix?












0












$begingroup$


I have
$m$ of column vectors in $x_i in mathbb{R}^{ntimes 1}$ as
$$mathbf{X} = begin{bmatrix}x_1 & x_2 & ldots x_m end{bmatrix},$$



Then I multiply with a diagonal matrix, note that each element does not have to be equal, other wise a scalar matrix is too easy,
$$mathbf{Lambda} = begin{bmatrix}lambda_1 & & \
& ddots & \
& & lambda_nend{bmatrix}$$



What is the geometric meaning of $mathbf{X^prime} = mathbf{Lambda X}$?



What is the geometric relationship between $mathbf{X}^prime$ and $mathbf{X}$?



It seems to be pretty difficult to describe its behavior. It does not preserve the anlge between column vector in $mathbf{X}$.






linear-algebra matrices linear-transformations







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I have
    $m$ of column vectors in $x_i in mathbb{R}^{ntimes 1}$ as
    $$mathbf{X} = begin{bmatrix}x_1 & x_2 & ldots x_m end{bmatrix},$$



    Then I multiply with a diagonal matrix, note that each element does not have to be equal, other wise a scalar matrix is too easy,
    $$mathbf{Lambda} = begin{bmatrix}lambda_1 & & \
    & ddots & \
    & & lambda_nend{bmatrix}$$



    What is the geometric meaning of $mathbf{X^prime} = mathbf{Lambda X}$?



    What is the geometric relationship between $mathbf{X}^prime$ and $mathbf{X}$?



    It seems to be pretty difficult to describe its behavior. It does not preserve the anlge between column vector in $mathbf{X}$.






    linear-algebra matrices linear-transformations







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have
      $m$ of column vectors in $x_i in mathbb{R}^{ntimes 1}$ as
      $$mathbf{X} = begin{bmatrix}x_1 & x_2 & ldots x_m end{bmatrix},$$



      Then I multiply with a diagonal matrix, note that each element does not have to be equal, other wise a scalar matrix is too easy,
      $$mathbf{Lambda} = begin{bmatrix}lambda_1 & & \
      & ddots & \
      & & lambda_nend{bmatrix}$$



      What is the geometric meaning of $mathbf{X^prime} = mathbf{Lambda X}$?



      What is the geometric relationship between $mathbf{X}^prime$ and $mathbf{X}$?



      It seems to be pretty difficult to describe its behavior. It does not preserve the anlge between column vector in $mathbf{X}$.






      linear-algebra matrices linear-transformations







      share|cite|improve this question











      $endgroup$




      I have
      $m$ of column vectors in $x_i in mathbb{R}^{ntimes 1}$ as
      $$mathbf{X} = begin{bmatrix}x_1 & x_2 & ldots x_m end{bmatrix},$$



      Then I multiply with a diagonal matrix, note that each element does not have to be equal, other wise a scalar matrix is too easy,
      $$mathbf{Lambda} = begin{bmatrix}lambda_1 & & \
      & ddots & \
      & & lambda_nend{bmatrix}$$



      What is the geometric meaning of $mathbf{X^prime} = mathbf{Lambda X}$?



      What is the geometric relationship between $mathbf{X}^prime$ and $mathbf{X}$?



      It seems to be pretty difficult to describe its behavior. It does not preserve the anlge between column vector in $mathbf{X}$.






      linear-algebra matrices linear-transformations




      linear-algebra matrices linear-transformations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 2:33







      ArtificiallyIntelligence

















      asked Jan 7 at 2:19









      ArtificiallyIntelligenceArtificiallyIntelligence

      310112




      310112






















          2 Answers
          2






          active

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          1












          $begingroup$

          Suppose that all $lambda_i$ are non-zero. One way to think of this action is to consider the fact that
          $$
          Lambda X = (Lambda X Lambda^{-1}) (Lambda LambdaLambda^{-1}) = [X]_{mathcal B} [Lambda]_{mathcal B}
          $$
          The transition $X mapsto Lambda X Lambda^{-1}$ can be thought of geometrically as a change in basis. That is, we reinterpret $X$ as a transformation relative to the basis $mathcal B = {lambda_1 e_1, dots, lambda_n e_n}$, where $e_i$ is the $i$th standard basis vector.



          Multiplying from the right by $Lambda$, as before, amounts to scaling the columns.




          Or, if you prefer: multiply first, then change basis, since
          $$
          Lambda X = Lambda [X Lambda]Lambda^{-1}
          $$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Notice that when you multiple $Lambda$ to X, the resultant matrix is $left[ lambda_1 x_1,,,,,lambda_2 x_2,,,ldots lambda_n x_nright]$. You multiply each column but the corresponding scalar.



            The relationship between X$^prime$ and X is that it stretches (or contracts) each column vector.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
              $endgroup$
              – ArtificiallyIntelligence
              Jan 7 at 2:32












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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Suppose that all $lambda_i$ are non-zero. One way to think of this action is to consider the fact that
            $$
            Lambda X = (Lambda X Lambda^{-1}) (Lambda LambdaLambda^{-1}) = [X]_{mathcal B} [Lambda]_{mathcal B}
            $$
            The transition $X mapsto Lambda X Lambda^{-1}$ can be thought of geometrically as a change in basis. That is, we reinterpret $X$ as a transformation relative to the basis $mathcal B = {lambda_1 e_1, dots, lambda_n e_n}$, where $e_i$ is the $i$th standard basis vector.



            Multiplying from the right by $Lambda$, as before, amounts to scaling the columns.




            Or, if you prefer: multiply first, then change basis, since
            $$
            Lambda X = Lambda [X Lambda]Lambda^{-1}
            $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Suppose that all $lambda_i$ are non-zero. One way to think of this action is to consider the fact that
              $$
              Lambda X = (Lambda X Lambda^{-1}) (Lambda LambdaLambda^{-1}) = [X]_{mathcal B} [Lambda]_{mathcal B}
              $$
              The transition $X mapsto Lambda X Lambda^{-1}$ can be thought of geometrically as a change in basis. That is, we reinterpret $X$ as a transformation relative to the basis $mathcal B = {lambda_1 e_1, dots, lambda_n e_n}$, where $e_i$ is the $i$th standard basis vector.



              Multiplying from the right by $Lambda$, as before, amounts to scaling the columns.




              Or, if you prefer: multiply first, then change basis, since
              $$
              Lambda X = Lambda [X Lambda]Lambda^{-1}
              $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Suppose that all $lambda_i$ are non-zero. One way to think of this action is to consider the fact that
                $$
                Lambda X = (Lambda X Lambda^{-1}) (Lambda LambdaLambda^{-1}) = [X]_{mathcal B} [Lambda]_{mathcal B}
                $$
                The transition $X mapsto Lambda X Lambda^{-1}$ can be thought of geometrically as a change in basis. That is, we reinterpret $X$ as a transformation relative to the basis $mathcal B = {lambda_1 e_1, dots, lambda_n e_n}$, where $e_i$ is the $i$th standard basis vector.



                Multiplying from the right by $Lambda$, as before, amounts to scaling the columns.




                Or, if you prefer: multiply first, then change basis, since
                $$
                Lambda X = Lambda [X Lambda]Lambda^{-1}
                $$






                share|cite|improve this answer









                $endgroup$



                Suppose that all $lambda_i$ are non-zero. One way to think of this action is to consider the fact that
                $$
                Lambda X = (Lambda X Lambda^{-1}) (Lambda LambdaLambda^{-1}) = [X]_{mathcal B} [Lambda]_{mathcal B}
                $$
                The transition $X mapsto Lambda X Lambda^{-1}$ can be thought of geometrically as a change in basis. That is, we reinterpret $X$ as a transformation relative to the basis $mathcal B = {lambda_1 e_1, dots, lambda_n e_n}$, where $e_i$ is the $i$th standard basis vector.



                Multiplying from the right by $Lambda$, as before, amounts to scaling the columns.




                Or, if you prefer: multiply first, then change basis, since
                $$
                Lambda X = Lambda [X Lambda]Lambda^{-1}
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 7 at 3:06









                OmnomnomnomOmnomnomnom

                129k794188




                129k794188























                    0












                    $begingroup$

                    Notice that when you multiple $Lambda$ to X, the resultant matrix is $left[ lambda_1 x_1,,,,,lambda_2 x_2,,,ldots lambda_n x_nright]$. You multiply each column but the corresponding scalar.



                    The relationship between X$^prime$ and X is that it stretches (or contracts) each column vector.






                    share|cite|improve this answer









                    $endgroup$









                    • 1




                      $begingroup$
                      Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
                      $endgroup$
                      – ArtificiallyIntelligence
                      Jan 7 at 2:32
















                    0












                    $begingroup$

                    Notice that when you multiple $Lambda$ to X, the resultant matrix is $left[ lambda_1 x_1,,,,,lambda_2 x_2,,,ldots lambda_n x_nright]$. You multiply each column but the corresponding scalar.



                    The relationship between X$^prime$ and X is that it stretches (or contracts) each column vector.






                    share|cite|improve this answer









                    $endgroup$









                    • 1




                      $begingroup$
                      Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
                      $endgroup$
                      – ArtificiallyIntelligence
                      Jan 7 at 2:32














                    0












                    0








                    0





                    $begingroup$

                    Notice that when you multiple $Lambda$ to X, the resultant matrix is $left[ lambda_1 x_1,,,,,lambda_2 x_2,,,ldots lambda_n x_nright]$. You multiply each column but the corresponding scalar.



                    The relationship between X$^prime$ and X is that it stretches (or contracts) each column vector.






                    share|cite|improve this answer









                    $endgroup$



                    Notice that when you multiple $Lambda$ to X, the resultant matrix is $left[ lambda_1 x_1,,,,,lambda_2 x_2,,,ldots lambda_n x_nright]$. You multiply each column but the corresponding scalar.



                    The relationship between X$^prime$ and X is that it stretches (or contracts) each column vector.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 7 at 2:29









                    Joel PereiraJoel Pereira

                    83719




                    83719








                    • 1




                      $begingroup$
                      Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
                      $endgroup$
                      – ArtificiallyIntelligence
                      Jan 7 at 2:32














                    • 1




                      $begingroup$
                      Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
                      $endgroup$
                      – ArtificiallyIntelligence
                      Jan 7 at 2:32








                    1




                    1




                    $begingroup$
                    Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
                    $endgroup$
                    – ArtificiallyIntelligence
                    Jan 7 at 2:32




                    $begingroup$
                    Thank you for your reply. But isn't that a right multiplication of the diagonal matrix?
                    $endgroup$
                    – ArtificiallyIntelligence
                    Jan 7 at 2:32


















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