Help with the question if the following is a linear transformation or not












0












$begingroup$


I am new to Linear Algebra, and am asked to check if the following transformation is linear. In principle, I know the criteria to check for: T(v+v')=T(v)+T(v') and c(Tv)=T(cv), where v is a vector, c is any scalar, and T is the transformation. However, when applying these to the following transformation, I get completely lost because of the complexity of the actual transformation. I am thinking there must be a shortcut I am not seeing. Any help would be very much appreciated.



$T:M^C_{nxn}to M^C_{nxn}$ defined by $T(X)=2X+(1+3i)bar X^t$. $M^C_{nxn}$ is a linear space over C.



There is an additional comment stating that we should refer to the general element of matrix $X$ as $x_{ij}$, while referring to the general element of matrix $bar X$ as $bar x_{ij}$



Thank you!






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try writing the statement and proof for $n=2$. With $2 times 2$ matrices you won't need any indices for the "general element". Then work on the general case. You can edit the question to show us how far you get before you're stuck.
    $endgroup$
    – Ethan Bolker
    Dec 31 '18 at 15:08










  • $begingroup$
    I fail to see why the subexpression $S: X mapsto bar{X}$ should be linear. $S(lambda X) = bar{lambda} S(X)$, so it isn't linear. Anyway, for OP it's probably easier to prove linearity of $Xmapsto 2X$, $S$, $Xmapsto X^t$ and $Xmapsto (1+3i)X$ separately and then apply "addition of linear functions is linear".
    $endgroup$
    – ComFreek
    Dec 31 '18 at 15:16
















0












$begingroup$


I am new to Linear Algebra, and am asked to check if the following transformation is linear. In principle, I know the criteria to check for: T(v+v')=T(v)+T(v') and c(Tv)=T(cv), where v is a vector, c is any scalar, and T is the transformation. However, when applying these to the following transformation, I get completely lost because of the complexity of the actual transformation. I am thinking there must be a shortcut I am not seeing. Any help would be very much appreciated.



$T:M^C_{nxn}to M^C_{nxn}$ defined by $T(X)=2X+(1+3i)bar X^t$. $M^C_{nxn}$ is a linear space over C.



There is an additional comment stating that we should refer to the general element of matrix $X$ as $x_{ij}$, while referring to the general element of matrix $bar X$ as $bar x_{ij}$



Thank you!






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try writing the statement and proof for $n=2$. With $2 times 2$ matrices you won't need any indices for the "general element". Then work on the general case. You can edit the question to show us how far you get before you're stuck.
    $endgroup$
    – Ethan Bolker
    Dec 31 '18 at 15:08










  • $begingroup$
    I fail to see why the subexpression $S: X mapsto bar{X}$ should be linear. $S(lambda X) = bar{lambda} S(X)$, so it isn't linear. Anyway, for OP it's probably easier to prove linearity of $Xmapsto 2X$, $S$, $Xmapsto X^t$ and $Xmapsto (1+3i)X$ separately and then apply "addition of linear functions is linear".
    $endgroup$
    – ComFreek
    Dec 31 '18 at 15:16














0












0








0


1



$begingroup$


I am new to Linear Algebra, and am asked to check if the following transformation is linear. In principle, I know the criteria to check for: T(v+v')=T(v)+T(v') and c(Tv)=T(cv), where v is a vector, c is any scalar, and T is the transformation. However, when applying these to the following transformation, I get completely lost because of the complexity of the actual transformation. I am thinking there must be a shortcut I am not seeing. Any help would be very much appreciated.



$T:M^C_{nxn}to M^C_{nxn}$ defined by $T(X)=2X+(1+3i)bar X^t$. $M^C_{nxn}$ is a linear space over C.



There is an additional comment stating that we should refer to the general element of matrix $X$ as $x_{ij}$, while referring to the general element of matrix $bar X$ as $bar x_{ij}$



Thank you!






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$




I am new to Linear Algebra, and am asked to check if the following transformation is linear. In principle, I know the criteria to check for: T(v+v')=T(v)+T(v') and c(Tv)=T(cv), where v is a vector, c is any scalar, and T is the transformation. However, when applying these to the following transformation, I get completely lost because of the complexity of the actual transformation. I am thinking there must be a shortcut I am not seeing. Any help would be very much appreciated.



$T:M^C_{nxn}to M^C_{nxn}$ defined by $T(X)=2X+(1+3i)bar X^t$. $M^C_{nxn}$ is a linear space over C.



There is an additional comment stating that we should refer to the general element of matrix $X$ as $x_{ij}$, while referring to the general element of matrix $bar X$ as $bar x_{ij}$



Thank you!






linear-algebra linear-transformations




linear-algebra linear-transformations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 31 '18 at 15:04









daltadalta

11311




11311












  • $begingroup$
    Try writing the statement and proof for $n=2$. With $2 times 2$ matrices you won't need any indices for the "general element". Then work on the general case. You can edit the question to show us how far you get before you're stuck.
    $endgroup$
    – Ethan Bolker
    Dec 31 '18 at 15:08










  • $begingroup$
    I fail to see why the subexpression $S: X mapsto bar{X}$ should be linear. $S(lambda X) = bar{lambda} S(X)$, so it isn't linear. Anyway, for OP it's probably easier to prove linearity of $Xmapsto 2X$, $S$, $Xmapsto X^t$ and $Xmapsto (1+3i)X$ separately and then apply "addition of linear functions is linear".
    $endgroup$
    – ComFreek
    Dec 31 '18 at 15:16


















  • $begingroup$
    Try writing the statement and proof for $n=2$. With $2 times 2$ matrices you won't need any indices for the "general element". Then work on the general case. You can edit the question to show us how far you get before you're stuck.
    $endgroup$
    – Ethan Bolker
    Dec 31 '18 at 15:08










  • $begingroup$
    I fail to see why the subexpression $S: X mapsto bar{X}$ should be linear. $S(lambda X) = bar{lambda} S(X)$, so it isn't linear. Anyway, for OP it's probably easier to prove linearity of $Xmapsto 2X$, $S$, $Xmapsto X^t$ and $Xmapsto (1+3i)X$ separately and then apply "addition of linear functions is linear".
    $endgroup$
    – ComFreek
    Dec 31 '18 at 15:16
















$begingroup$
Try writing the statement and proof for $n=2$. With $2 times 2$ matrices you won't need any indices for the "general element". Then work on the general case. You can edit the question to show us how far you get before you're stuck.
$endgroup$
– Ethan Bolker
Dec 31 '18 at 15:08




$begingroup$
Try writing the statement and proof for $n=2$. With $2 times 2$ matrices you won't need any indices for the "general element". Then work on the general case. You can edit the question to show us how far you get before you're stuck.
$endgroup$
– Ethan Bolker
Dec 31 '18 at 15:08












$begingroup$
I fail to see why the subexpression $S: X mapsto bar{X}$ should be linear. $S(lambda X) = bar{lambda} S(X)$, so it isn't linear. Anyway, for OP it's probably easier to prove linearity of $Xmapsto 2X$, $S$, $Xmapsto X^t$ and $Xmapsto (1+3i)X$ separately and then apply "addition of linear functions is linear".
$endgroup$
– ComFreek
Dec 31 '18 at 15:16




$begingroup$
I fail to see why the subexpression $S: X mapsto bar{X}$ should be linear. $S(lambda X) = bar{lambda} S(X)$, so it isn't linear. Anyway, for OP it's probably easier to prove linearity of $Xmapsto 2X$, $S$, $Xmapsto X^t$ and $Xmapsto (1+3i)X$ separately and then apply "addition of linear functions is linear".
$endgroup$
– ComFreek
Dec 31 '18 at 15:16










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint:



Apply the rules mechanically:



$$2(X+X')+(1+3i)overline{(X+X')}^t
\=2X+2X'+(1+3i)(bar X+bar X')^t
\=2X+2X'+(1+3i)(bar X^t+bar X'^t)
\=2X+2X'+(1+3i)bar X+(1+3i)bar X'
\=(2X+(1+3i)bar X)+(2X'+(1+3i)bar X').$$



There is nothing really complicated.



You may also just state that




  • averaging is a linear operation,

  • transposition is a linear operation,

  • scalar multiplicationis a linear operation,

  • addition is a linear operation.


Proving each of theses statements separately is at your reach.




Update:



I didn't see that the overline denotes conjugation. Then




  • complex conjugation is not is a linear operation.


Because $$overline{aX}=overline aoverline Xne aoverline X.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But its not linear right?
    $endgroup$
    – Olof Rubin
    Dec 31 '18 at 15:20










  • $begingroup$
    @OlofRubin: why not ? I just explained that it is.
    $endgroup$
    – Yves Daoust
    Dec 31 '18 at 15:21










  • $begingroup$
    $T(iX)neq iT(X)$ in general?
    $endgroup$
    – Olof Rubin
    Dec 31 '18 at 15:22










  • $begingroup$
    @OlofRubin: how so ? Please substantitate.
    $endgroup$
    – Yves Daoust
    Dec 31 '18 at 15:22












  • $begingroup$
    $T(iI) = 2iI-i(1+3i)I = iI+3I$ while $iT(I) = 2iI+i(1+3i)I = 3iI-3I$?
    $endgroup$
    – Olof Rubin
    Dec 31 '18 at 15:23





















1












$begingroup$

The transformation clearly satisfies $T(X+Y) = T(X)+T(Y)$ right? Also if $alpha$ is real then $T(alpha X) = alpha T(X)$ is trivial. However the final property we need for linearity is that $T(iX) = iT(X)$ for then we can conclude that $T((alpha+ibeta)X) = alpha T(X)+ibeta T(X)$ for real $alpha,beta$. We therefore consider



$$2iX+i(1+3i)bar{X}^t=i(2X+(1+3i)bar{X}^t) = iT(X) =^? T(iX) = 2iX-i(1+3i)bar{X}^t$$



If we consider the identity matrix $I$ then is it true that



$$2iI+i(1+3i)I =^? 2iI-i(1+3i)I$$






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint:



    Apply the rules mechanically:



    $$2(X+X')+(1+3i)overline{(X+X')}^t
    \=2X+2X'+(1+3i)(bar X+bar X')^t
    \=2X+2X'+(1+3i)(bar X^t+bar X'^t)
    \=2X+2X'+(1+3i)bar X+(1+3i)bar X'
    \=(2X+(1+3i)bar X)+(2X'+(1+3i)bar X').$$



    There is nothing really complicated.



    You may also just state that




    • averaging is a linear operation,

    • transposition is a linear operation,

    • scalar multiplicationis a linear operation,

    • addition is a linear operation.


    Proving each of theses statements separately is at your reach.




    Update:



    I didn't see that the overline denotes conjugation. Then




    • complex conjugation is not is a linear operation.


    Because $$overline{aX}=overline aoverline Xne aoverline X.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But its not linear right?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:20










    • $begingroup$
      @OlofRubin: why not ? I just explained that it is.
      $endgroup$
      – Yves Daoust
      Dec 31 '18 at 15:21










    • $begingroup$
      $T(iX)neq iT(X)$ in general?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:22










    • $begingroup$
      @OlofRubin: how so ? Please substantitate.
      $endgroup$
      – Yves Daoust
      Dec 31 '18 at 15:22












    • $begingroup$
      $T(iI) = 2iI-i(1+3i)I = iI+3I$ while $iT(I) = 2iI+i(1+3i)I = 3iI-3I$?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:23


















    1












    $begingroup$

    Hint:



    Apply the rules mechanically:



    $$2(X+X')+(1+3i)overline{(X+X')}^t
    \=2X+2X'+(1+3i)(bar X+bar X')^t
    \=2X+2X'+(1+3i)(bar X^t+bar X'^t)
    \=2X+2X'+(1+3i)bar X+(1+3i)bar X'
    \=(2X+(1+3i)bar X)+(2X'+(1+3i)bar X').$$



    There is nothing really complicated.



    You may also just state that




    • averaging is a linear operation,

    • transposition is a linear operation,

    • scalar multiplicationis a linear operation,

    • addition is a linear operation.


    Proving each of theses statements separately is at your reach.




    Update:



    I didn't see that the overline denotes conjugation. Then




    • complex conjugation is not is a linear operation.


    Because $$overline{aX}=overline aoverline Xne aoverline X.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But its not linear right?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:20










    • $begingroup$
      @OlofRubin: why not ? I just explained that it is.
      $endgroup$
      – Yves Daoust
      Dec 31 '18 at 15:21










    • $begingroup$
      $T(iX)neq iT(X)$ in general?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:22










    • $begingroup$
      @OlofRubin: how so ? Please substantitate.
      $endgroup$
      – Yves Daoust
      Dec 31 '18 at 15:22












    • $begingroup$
      $T(iI) = 2iI-i(1+3i)I = iI+3I$ while $iT(I) = 2iI+i(1+3i)I = 3iI-3I$?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:23
















    1












    1








    1





    $begingroup$

    Hint:



    Apply the rules mechanically:



    $$2(X+X')+(1+3i)overline{(X+X')}^t
    \=2X+2X'+(1+3i)(bar X+bar X')^t
    \=2X+2X'+(1+3i)(bar X^t+bar X'^t)
    \=2X+2X'+(1+3i)bar X+(1+3i)bar X'
    \=(2X+(1+3i)bar X)+(2X'+(1+3i)bar X').$$



    There is nothing really complicated.



    You may also just state that




    • averaging is a linear operation,

    • transposition is a linear operation,

    • scalar multiplicationis a linear operation,

    • addition is a linear operation.


    Proving each of theses statements separately is at your reach.




    Update:



    I didn't see that the overline denotes conjugation. Then




    • complex conjugation is not is a linear operation.


    Because $$overline{aX}=overline aoverline Xne aoverline X.$$






    share|cite|improve this answer











    $endgroup$



    Hint:



    Apply the rules mechanically:



    $$2(X+X')+(1+3i)overline{(X+X')}^t
    \=2X+2X'+(1+3i)(bar X+bar X')^t
    \=2X+2X'+(1+3i)(bar X^t+bar X'^t)
    \=2X+2X'+(1+3i)bar X+(1+3i)bar X'
    \=(2X+(1+3i)bar X)+(2X'+(1+3i)bar X').$$



    There is nothing really complicated.



    You may also just state that




    • averaging is a linear operation,

    • transposition is a linear operation,

    • scalar multiplicationis a linear operation,

    • addition is a linear operation.


    Proving each of theses statements separately is at your reach.




    Update:



    I didn't see that the overline denotes conjugation. Then




    • complex conjugation is not is a linear operation.


    Because $$overline{aX}=overline aoverline Xne aoverline X.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 31 '18 at 15:25

























    answered Dec 31 '18 at 15:17









    Yves DaoustYves Daoust

    134k676232




    134k676232












    • $begingroup$
      But its not linear right?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:20










    • $begingroup$
      @OlofRubin: why not ? I just explained that it is.
      $endgroup$
      – Yves Daoust
      Dec 31 '18 at 15:21










    • $begingroup$
      $T(iX)neq iT(X)$ in general?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:22










    • $begingroup$
      @OlofRubin: how so ? Please substantitate.
      $endgroup$
      – Yves Daoust
      Dec 31 '18 at 15:22












    • $begingroup$
      $T(iI) = 2iI-i(1+3i)I = iI+3I$ while $iT(I) = 2iI+i(1+3i)I = 3iI-3I$?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:23




















    • $begingroup$
      But its not linear right?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:20










    • $begingroup$
      @OlofRubin: why not ? I just explained that it is.
      $endgroup$
      – Yves Daoust
      Dec 31 '18 at 15:21










    • $begingroup$
      $T(iX)neq iT(X)$ in general?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:22










    • $begingroup$
      @OlofRubin: how so ? Please substantitate.
      $endgroup$
      – Yves Daoust
      Dec 31 '18 at 15:22












    • $begingroup$
      $T(iI) = 2iI-i(1+3i)I = iI+3I$ while $iT(I) = 2iI+i(1+3i)I = 3iI-3I$?
      $endgroup$
      – Olof Rubin
      Dec 31 '18 at 15:23


















    $begingroup$
    But its not linear right?
    $endgroup$
    – Olof Rubin
    Dec 31 '18 at 15:20




    $begingroup$
    But its not linear right?
    $endgroup$
    – Olof Rubin
    Dec 31 '18 at 15:20












    $begingroup$
    @OlofRubin: why not ? I just explained that it is.
    $endgroup$
    – Yves Daoust
    Dec 31 '18 at 15:21




    $begingroup$
    @OlofRubin: why not ? I just explained that it is.
    $endgroup$
    – Yves Daoust
    Dec 31 '18 at 15:21












    $begingroup$
    $T(iX)neq iT(X)$ in general?
    $endgroup$
    – Olof Rubin
    Dec 31 '18 at 15:22




    $begingroup$
    $T(iX)neq iT(X)$ in general?
    $endgroup$
    – Olof Rubin
    Dec 31 '18 at 15:22












    $begingroup$
    @OlofRubin: how so ? Please substantitate.
    $endgroup$
    – Yves Daoust
    Dec 31 '18 at 15:22






    $begingroup$
    @OlofRubin: how so ? Please substantitate.
    $endgroup$
    – Yves Daoust
    Dec 31 '18 at 15:22














    $begingroup$
    $T(iI) = 2iI-i(1+3i)I = iI+3I$ while $iT(I) = 2iI+i(1+3i)I = 3iI-3I$?
    $endgroup$
    – Olof Rubin
    Dec 31 '18 at 15:23






    $begingroup$
    $T(iI) = 2iI-i(1+3i)I = iI+3I$ while $iT(I) = 2iI+i(1+3i)I = 3iI-3I$?
    $endgroup$
    – Olof Rubin
    Dec 31 '18 at 15:23













    1












    $begingroup$

    The transformation clearly satisfies $T(X+Y) = T(X)+T(Y)$ right? Also if $alpha$ is real then $T(alpha X) = alpha T(X)$ is trivial. However the final property we need for linearity is that $T(iX) = iT(X)$ for then we can conclude that $T((alpha+ibeta)X) = alpha T(X)+ibeta T(X)$ for real $alpha,beta$. We therefore consider



    $$2iX+i(1+3i)bar{X}^t=i(2X+(1+3i)bar{X}^t) = iT(X) =^? T(iX) = 2iX-i(1+3i)bar{X}^t$$



    If we consider the identity matrix $I$ then is it true that



    $$2iI+i(1+3i)I =^? 2iI-i(1+3i)I$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The transformation clearly satisfies $T(X+Y) = T(X)+T(Y)$ right? Also if $alpha$ is real then $T(alpha X) = alpha T(X)$ is trivial. However the final property we need for linearity is that $T(iX) = iT(X)$ for then we can conclude that $T((alpha+ibeta)X) = alpha T(X)+ibeta T(X)$ for real $alpha,beta$. We therefore consider



      $$2iX+i(1+3i)bar{X}^t=i(2X+(1+3i)bar{X}^t) = iT(X) =^? T(iX) = 2iX-i(1+3i)bar{X}^t$$



      If we consider the identity matrix $I$ then is it true that



      $$2iI+i(1+3i)I =^? 2iI-i(1+3i)I$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The transformation clearly satisfies $T(X+Y) = T(X)+T(Y)$ right? Also if $alpha$ is real then $T(alpha X) = alpha T(X)$ is trivial. However the final property we need for linearity is that $T(iX) = iT(X)$ for then we can conclude that $T((alpha+ibeta)X) = alpha T(X)+ibeta T(X)$ for real $alpha,beta$. We therefore consider



        $$2iX+i(1+3i)bar{X}^t=i(2X+(1+3i)bar{X}^t) = iT(X) =^? T(iX) = 2iX-i(1+3i)bar{X}^t$$



        If we consider the identity matrix $I$ then is it true that



        $$2iI+i(1+3i)I =^? 2iI-i(1+3i)I$$






        share|cite|improve this answer









        $endgroup$



        The transformation clearly satisfies $T(X+Y) = T(X)+T(Y)$ right? Also if $alpha$ is real then $T(alpha X) = alpha T(X)$ is trivial. However the final property we need for linearity is that $T(iX) = iT(X)$ for then we can conclude that $T((alpha+ibeta)X) = alpha T(X)+ibeta T(X)$ for real $alpha,beta$. We therefore consider



        $$2iX+i(1+3i)bar{X}^t=i(2X+(1+3i)bar{X}^t) = iT(X) =^? T(iX) = 2iX-i(1+3i)bar{X}^t$$



        If we consider the identity matrix $I$ then is it true that



        $$2iI+i(1+3i)I =^? 2iI-i(1+3i)I$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 15:14









        Olof RubinOlof Rubin

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