mistake in calculations for ImT and KerT

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$begingroup$


I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?



The linear transformation $T:M_2x2(R) to R_3[x]$ is defined by:
$$Tbeginpmatrixa&b\c&d endpmatrix$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$beginpmatrixa&b\c&d endpmatrix$$$in M_2x2(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$beginpmatrixa&b\-b&a endpmatrix$$=$$abeginpmatrix1&0\0&1 endpmatrix$$+$$bbeginpmatrix0&1\-1&0 endpmatrix$$



So KerT = Sp$$beginpmatrix1&0\0&1 endpmatrix$$,$$beginpmatrix0&1\-1&0 endpmatrix$$
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.



For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.



The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_2x2(R)=4$










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  • $begingroup$
    Use $ X$ for $ X$.
    $endgroup$
    – Shaun
    Jan 3 at 12:38















3












$begingroup$


I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?



The linear transformation $T:M_2x2(R) to R_3[x]$ is defined by:
$$Tbeginpmatrixa&b\c&d endpmatrix$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$beginpmatrixa&b\c&d endpmatrix$$$in M_2x2(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$beginpmatrixa&b\-b&a endpmatrix$$=$$abeginpmatrix1&0\0&1 endpmatrix$$+$$bbeginpmatrix0&1\-1&0 endpmatrix$$



So KerT = Sp$$beginpmatrix1&0\0&1 endpmatrix$$,$$beginpmatrix0&1\-1&0 endpmatrix$$
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.



For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.



The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_2x2(R)=4$










share|cite|improve this question











$endgroup$











  • $begingroup$
    Use $ X$ for $ X$.
    $endgroup$
    – Shaun
    Jan 3 at 12:38













3












3








3





$begingroup$


I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?



The linear transformation $T:M_2x2(R) to R_3[x]$ is defined by:
$$Tbeginpmatrixa&b\c&d endpmatrix$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$beginpmatrixa&b\c&d endpmatrix$$$in M_2x2(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$beginpmatrixa&b\-b&a endpmatrix$$=$$abeginpmatrix1&0\0&1 endpmatrix$$+$$bbeginpmatrix0&1\-1&0 endpmatrix$$



So KerT = Sp$$beginpmatrix1&0\0&1 endpmatrix$$,$$beginpmatrix0&1\-1&0 endpmatrix$$
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.



For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.



The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_2x2(R)=4$










share|cite|improve this question











$endgroup$




I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?



The linear transformation $T:M_2x2(R) to R_3[x]$ is defined by:
$$Tbeginpmatrixa&b\c&d endpmatrix$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$beginpmatrixa&b\c&d endpmatrix$$$in M_2x2(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$beginpmatrixa&b\-b&a endpmatrix$$=$$abeginpmatrix1&0\0&1 endpmatrix$$+$$bbeginpmatrix0&1\-1&0 endpmatrix$$



So KerT = Sp$$beginpmatrix1&0\0&1 endpmatrix$$,$$beginpmatrix0&1\-1&0 endpmatrix$$
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.



For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.



The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_2x2(R)=4$







linear-algebra linear-transformations






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edited Jan 3 at 12:36







dalta

















asked Jan 3 at 12:07









daltadalta

627




627











  • $begingroup$
    Use $ X$ for $ X$.
    $endgroup$
    – Shaun
    Jan 3 at 12:38
















  • $begingroup$
    Use $ X$ for $ X$.
    $endgroup$
    – Shaun
    Jan 3 at 12:38















$begingroup$
Use $ X$ for $ X$.
$endgroup$
– Shaun
Jan 3 at 12:38




$begingroup$
Use $ X$ for $ X$.
$endgroup$
– Shaun
Jan 3 at 12:38










2 Answers
2






active

oldest

votes


















4












$begingroup$


For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.




It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
$$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...




Or for a more general approach, rewrite as (split in all the coefficients):
$$(a-d)x^2+(b+c)x+5a-5d = a(colorbluex^2+5)+bcolorredx+ccolorpurplex+d(colorgreen-x^2-5)$$
and reduce the spanning set $leftcolorbluex^2+5,colorredx,colorpurplex,colorgreen-x^2-5right$ to a basis by eliminating the linearly dependent elements.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Thank you! So my calculations for KerT were correct?
    $endgroup$
    – dalta
    Jan 3 at 12:34






  • 1




    $begingroup$
    Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
    $endgroup$
    – StackTD
    Jan 3 at 12:35



















2












$begingroup$

$$textIm(T)netextSp1,x,x^2$$



This is because $TBig(beginbmatrixa&b\c&d endbmatrixBig)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$textIm(T)=textSpx,x^2+5$$ which is also a basis with dimension $2$.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$


    For ImT:
    $(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.




    It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
    $$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
    so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...




    Or for a more general approach, rewrite as (split in all the coefficients):
    $$(a-d)x^2+(b+c)x+5a-5d = a(colorbluex^2+5)+bcolorredx+ccolorpurplex+d(colorgreen-x^2-5)$$
    and reduce the spanning set $leftcolorbluex^2+5,colorredx,colorpurplex,colorgreen-x^2-5right$ to a basis by eliminating the linearly dependent elements.






    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      Thank you! So my calculations for KerT were correct?
      $endgroup$
      – dalta
      Jan 3 at 12:34






    • 1




      $begingroup$
      Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
      $endgroup$
      – StackTD
      Jan 3 at 12:35
















    4












    $begingroup$


    For ImT:
    $(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.




    It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
    $$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
    so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...




    Or for a more general approach, rewrite as (split in all the coefficients):
    $$(a-d)x^2+(b+c)x+5a-5d = a(colorbluex^2+5)+bcolorredx+ccolorpurplex+d(colorgreen-x^2-5)$$
    and reduce the spanning set $leftcolorbluex^2+5,colorredx,colorpurplex,colorgreen-x^2-5right$ to a basis by eliminating the linearly dependent elements.






    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      Thank you! So my calculations for KerT were correct?
      $endgroup$
      – dalta
      Jan 3 at 12:34






    • 1




      $begingroup$
      Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
      $endgroup$
      – StackTD
      Jan 3 at 12:35














    4












    4








    4





    $begingroup$


    For ImT:
    $(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.




    It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
    $$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
    so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...




    Or for a more general approach, rewrite as (split in all the coefficients):
    $$(a-d)x^2+(b+c)x+5a-5d = a(colorbluex^2+5)+bcolorredx+ccolorpurplex+d(colorgreen-x^2-5)$$
    and reduce the spanning set $leftcolorbluex^2+5,colorredx,colorpurplex,colorgreen-x^2-5right$ to a basis by eliminating the linearly dependent elements.






    share|cite|improve this answer











    $endgroup$




    For ImT:
    $(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Spx^2,x,1$. This set is per definition linearly independent and therefore $Spx^2,x,1$ is a basis to ImT and its dimension is 3.




    It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
    $$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
    so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...




    Or for a more general approach, rewrite as (split in all the coefficients):
    $$(a-d)x^2+(b+c)x+5a-5d = a(colorbluex^2+5)+bcolorredx+ccolorpurplex+d(colorgreen-x^2-5)$$
    and reduce the spanning set $leftcolorbluex^2+5,colorredx,colorpurplex,colorgreen-x^2-5right$ to a basis by eliminating the linearly dependent elements.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 3 at 12:23

























    answered Jan 3 at 12:18









    StackTDStackTD

    22.7k2049




    22.7k2049







    • 1




      $begingroup$
      Thank you! So my calculations for KerT were correct?
      $endgroup$
      – dalta
      Jan 3 at 12:34






    • 1




      $begingroup$
      Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
      $endgroup$
      – StackTD
      Jan 3 at 12:35













    • 1




      $begingroup$
      Thank you! So my calculations for KerT were correct?
      $endgroup$
      – dalta
      Jan 3 at 12:34






    • 1




      $begingroup$
      Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
      $endgroup$
      – StackTD
      Jan 3 at 12:35








    1




    1




    $begingroup$
    Thank you! So my calculations for KerT were correct?
    $endgroup$
    – dalta
    Jan 3 at 12:34




    $begingroup$
    Thank you! So my calculations for KerT were correct?
    $endgroup$
    – dalta
    Jan 3 at 12:34




    1




    1




    $begingroup$
    Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
    $endgroup$
    – StackTD
    Jan 3 at 12:35





    $begingroup$
    Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
    $endgroup$
    – StackTD
    Jan 3 at 12:35












    2












    $begingroup$

    $$textIm(T)netextSp1,x,x^2$$



    This is because $TBig(beginbmatrixa&b\c&d endbmatrixBig)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$textIm(T)=textSpx,x^2+5$$ which is also a basis with dimension $2$.






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      $$textIm(T)netextSp1,x,x^2$$



      This is because $TBig(beginbmatrixa&b\c&d endbmatrixBig)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$textIm(T)=textSpx,x^2+5$$ which is also a basis with dimension $2$.






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        $$textIm(T)netextSp1,x,x^2$$



        This is because $TBig(beginbmatrixa&b\c&d endbmatrixBig)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$textIm(T)=textSpx,x^2+5$$ which is also a basis with dimension $2$.






        share|cite|improve this answer









        $endgroup$



        $$textIm(T)netextSp1,x,x^2$$



        This is because $TBig(beginbmatrixa&b\c&d endbmatrixBig)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$textIm(T)=textSpx,x^2+5$$ which is also a basis with dimension $2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 12:23









        Shubham JohriShubham Johri

        4,769717




        4,769717



























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