mistake in calculations for ImT and KerT












3














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:
$$Tbegin{pmatrix}a&b\c&d end{pmatrix}$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$begin{pmatrix}a&b\c&d end{pmatrix}$$$in M_{2x2}(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$begin{pmatrix}a&b\-b&a end{pmatrix}$$=$$abegin{pmatrix}1&0\0&1 end{pmatrix}$$+$$bbegin{pmatrix}0&1\-1&0 end{pmatrix}$$



So KerT = Sp{$$begin{pmatrix}1&0\0&1 end{pmatrix}$$,$$begin{pmatrix}0&1\-1&0 end{pmatrix}$$}
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 $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ 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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  • Use ${ X}$ for ${ X}$.
    – Shaun
    1 hour ago
















3














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:
$$Tbegin{pmatrix}a&b\c&d end{pmatrix}$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$begin{pmatrix}a&b\c&d end{pmatrix}$$$in M_{2x2}(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$begin{pmatrix}a&b\-b&a end{pmatrix}$$=$$abegin{pmatrix}1&0\0&1 end{pmatrix}$$+$$bbegin{pmatrix}0&1\-1&0 end{pmatrix}$$



So KerT = Sp{$$begin{pmatrix}1&0\0&1 end{pmatrix}$$,$$begin{pmatrix}0&1\-1&0 end{pmatrix}$$}
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 $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ 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







share|cite|improve this question
























  • Use ${ X}$ for ${ X}$.
    – Shaun
    1 hour ago














3












3








3







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:
$$Tbegin{pmatrix}a&b\c&d end{pmatrix}$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$begin{pmatrix}a&b\c&d end{pmatrix}$$$in M_{2x2}(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$begin{pmatrix}a&b\-b&a end{pmatrix}$$=$$abegin{pmatrix}1&0\0&1 end{pmatrix}$$+$$bbegin{pmatrix}0&1\-1&0 end{pmatrix}$$



So KerT = Sp{$$begin{pmatrix}1&0\0&1 end{pmatrix}$$,$$begin{pmatrix}0&1\-1&0 end{pmatrix}$$}
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 $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ 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







share|cite|improve this question















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:
$$Tbegin{pmatrix}a&b\c&d end{pmatrix}$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$begin{pmatrix}a&b\c&d end{pmatrix}$$$in M_{2x2}(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$begin{pmatrix}a&b\-b&a end{pmatrix}$$=$$abegin{pmatrix}1&0\0&1 end{pmatrix}$$+$$bbegin{pmatrix}0&1\-1&0 end{pmatrix}$$



So KerT = Sp{$$begin{pmatrix}1&0\0&1 end{pmatrix}$$,$$begin{pmatrix}0&1\-1&0 end{pmatrix}$$}
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 $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ 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




linear-algebra linear-transformations






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share|cite|improve this question













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edited 1 hour ago

























asked 2 hours ago









dalta

577




577












  • Use ${ X}$ for ${ X}$.
    – Shaun
    1 hour ago


















  • Use ${ X}$ for ${ X}$.
    – Shaun
    1 hour ago
















Use ${ X}$ for ${ X}$.
– Shaun
1 hour ago




Use ${ X}$ for ${ X}$.
– Shaun
1 hour ago










2 Answers
2






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oldest

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4















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 $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ 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):
$$(a-d)x^2+(b+c)x+5a-5d = a(color{blue}{x^2+5})+bcolor{red}{x}+ccolor{purple}{x}+d(color{green}{-x^2-5})$$
and reduce the spanning set $left{color{blue}{x^2+5},color{red}{x},color{purple}{x},color{green}{-x^2-5}right}$ to a basis by eliminating the linearly dependent elements.






share|cite|improve this answer



















  • 1




    Thank you! So my calculations for KerT were correct?
    – dalta
    1 hour ago






  • 1




    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.
    – StackTD
    1 hour ago





















2














$$text{Im}(T)netext{Sp}{1,x,x^2}$$



This is because $TBig(begin{bmatrix}a&b\c&d end{bmatrix}Big)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$text{Im}(T)=text{Sp}{x,x^2+5}$$ which is also a basis with dimension $2$.






share|cite|improve this answer





















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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    4















    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 $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ 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):
    $$(a-d)x^2+(b+c)x+5a-5d = a(color{blue}{x^2+5})+bcolor{red}{x}+ccolor{purple}{x}+d(color{green}{-x^2-5})$$
    and reduce the spanning set $left{color{blue}{x^2+5},color{red}{x},color{purple}{x},color{green}{-x^2-5}right}$ to a basis by eliminating the linearly dependent elements.






    share|cite|improve this answer



















    • 1




      Thank you! So my calculations for KerT were correct?
      – dalta
      1 hour ago






    • 1




      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.
      – StackTD
      1 hour ago


















    4















    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 $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ 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):
    $$(a-d)x^2+(b+c)x+5a-5d = a(color{blue}{x^2+5})+bcolor{red}{x}+ccolor{purple}{x}+d(color{green}{-x^2-5})$$
    and reduce the spanning set $left{color{blue}{x^2+5},color{red}{x},color{purple}{x},color{green}{-x^2-5}right}$ to a basis by eliminating the linearly dependent elements.






    share|cite|improve this answer



















    • 1




      Thank you! So my calculations for KerT were correct?
      – dalta
      1 hour ago






    • 1




      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.
      – StackTD
      1 hour ago
















    4












    4








    4







    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 $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ 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):
    $$(a-d)x^2+(b+c)x+5a-5d = a(color{blue}{x^2+5})+bcolor{red}{x}+ccolor{purple}{x}+d(color{green}{-x^2-5})$$
    and reduce the spanning set $left{color{blue}{x^2+5},color{red}{x},color{purple}{x},color{green}{-x^2-5}right}$ to a basis by eliminating the linearly dependent elements.






    share|cite|improve this answer















    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 $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ 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):
    $$(a-d)x^2+(b+c)x+5a-5d = a(color{blue}{x^2+5})+bcolor{red}{x}+ccolor{purple}{x}+d(color{green}{-x^2-5})$$
    and reduce the spanning set $left{color{blue}{x^2+5},color{red}{x},color{purple}{x},color{green}{-x^2-5}right}$ 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 1 hour ago

























    answered 2 hours ago









    StackTD

    22.2k1947




    22.2k1947








    • 1




      Thank you! So my calculations for KerT were correct?
      – dalta
      1 hour ago






    • 1




      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.
      – StackTD
      1 hour ago
















    • 1




      Thank you! So my calculations for KerT were correct?
      – dalta
      1 hour ago






    • 1




      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.
      – StackTD
      1 hour ago










    1




    1




    Thank you! So my calculations for KerT were correct?
    – dalta
    1 hour ago




    Thank you! So my calculations for KerT were correct?
    – dalta
    1 hour ago




    1




    1




    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.
    – StackTD
    1 hour ago






    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.
    – StackTD
    1 hour ago













    2














    $$text{Im}(T)netext{Sp}{1,x,x^2}$$



    This is because $TBig(begin{bmatrix}a&b\c&d end{bmatrix}Big)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$text{Im}(T)=text{Sp}{x,x^2+5}$$ which is also a basis with dimension $2$.






    share|cite|improve this answer


























      2














      $$text{Im}(T)netext{Sp}{1,x,x^2}$$



      This is because $TBig(begin{bmatrix}a&b\c&d end{bmatrix}Big)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$text{Im}(T)=text{Sp}{x,x^2+5}$$ which is also a basis with dimension $2$.






      share|cite|improve this answer
























        2












        2








        2






        $$text{Im}(T)netext{Sp}{1,x,x^2}$$



        This is because $TBig(begin{bmatrix}a&b\c&d end{bmatrix}Big)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$text{Im}(T)=text{Sp}{x,x^2+5}$$ which is also a basis with dimension $2$.






        share|cite|improve this answer












        $$text{Im}(T)netext{Sp}{1,x,x^2}$$



        This is because $TBig(begin{bmatrix}a&b\c&d end{bmatrix}Big)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$text{Im}(T)=text{Sp}{x,x^2+5}$$ which is also a basis with dimension $2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Shubham Johri

        3,961717




        3,961717






























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