Linear Transformation involving polynomial and matrix












2














Determine if its linear the transformation



$f:Re_{3} [x]rightarrow M_{2x2}$



such as



$f(ax^3+bx^2+cx+d) = begin{bmatrix}
a-c & 0 \
0&b+d
end{bmatrix}$



for any $ax^3+bx^2+cx+d in Re _{3} [x]$



Im having trouble associating the polynomial with the resulting matrix.
I know I'm supposed to check the two main Linear transformation conditions but I don't know how to aproach this example.



Here's my attempt:



Condition 1: $T(u+v)=T(u)+T(v)$



$f((ax^3+bx^2+cx+d)+(a'x^3+b'x^2+c'x+d'))=f((ax^3+a'x^3)+(bx^2+b'x^2)+(cx+c'x)+(d+d'))=f((a+a')x^3+(b+b')x^2+(c+c')x+(d+d'))=begin{bmatrix}
(a+a')-(c+c') & 0\
0 & (b+b')+(d+d')
end{bmatrix} = begin{bmatrix}
a-c & 0\
0 & b+d
end{bmatrix} + begin{bmatrix}
a'-c' & 0\
0 & b'+d'
end{bmatrix}=f(ax^3+bx^2+cx+d)+f(a'x^3+b'x^2+c'x+d')$



Is this correct?






linear-transformations







share|cite|improve this question


















  • 3




    Looks right to me.
    – saulspatz
    2 days ago






  • 3




    Seems fine. ${}{}{}{} $
    – Thomas Shelby
    2 days ago
















2














Determine if its linear the transformation



$f:Re_{3} [x]rightarrow M_{2x2}$



such as



$f(ax^3+bx^2+cx+d) = begin{bmatrix}
a-c & 0 \
0&b+d
end{bmatrix}$



for any $ax^3+bx^2+cx+d in Re _{3} [x]$



Im having trouble associating the polynomial with the resulting matrix.
I know I'm supposed to check the two main Linear transformation conditions but I don't know how to aproach this example.



Here's my attempt:



Condition 1: $T(u+v)=T(u)+T(v)$



$f((ax^3+bx^2+cx+d)+(a'x^3+b'x^2+c'x+d'))=f((ax^3+a'x^3)+(bx^2+b'x^2)+(cx+c'x)+(d+d'))=f((a+a')x^3+(b+b')x^2+(c+c')x+(d+d'))=begin{bmatrix}
(a+a')-(c+c') & 0\
0 & (b+b')+(d+d')
end{bmatrix} = begin{bmatrix}
a-c & 0\
0 & b+d
end{bmatrix} + begin{bmatrix}
a'-c' & 0\
0 & b'+d'
end{bmatrix}=f(ax^3+bx^2+cx+d)+f(a'x^3+b'x^2+c'x+d')$



Is this correct?






linear-transformations







share|cite|improve this question


















  • 3




    Looks right to me.
    – saulspatz
    2 days ago






  • 3




    Seems fine. ${}{}{}{} $
    – Thomas Shelby
    2 days ago














2












2








2







Determine if its linear the transformation



$f:Re_{3} [x]rightarrow M_{2x2}$



such as



$f(ax^3+bx^2+cx+d) = begin{bmatrix}
a-c & 0 \
0&b+d
end{bmatrix}$



for any $ax^3+bx^2+cx+d in Re _{3} [x]$



Im having trouble associating the polynomial with the resulting matrix.
I know I'm supposed to check the two main Linear transformation conditions but I don't know how to aproach this example.



Here's my attempt:



Condition 1: $T(u+v)=T(u)+T(v)$



$f((ax^3+bx^2+cx+d)+(a'x^3+b'x^2+c'x+d'))=f((ax^3+a'x^3)+(bx^2+b'x^2)+(cx+c'x)+(d+d'))=f((a+a')x^3+(b+b')x^2+(c+c')x+(d+d'))=begin{bmatrix}
(a+a')-(c+c') & 0\
0 & (b+b')+(d+d')
end{bmatrix} = begin{bmatrix}
a-c & 0\
0 & b+d
end{bmatrix} + begin{bmatrix}
a'-c' & 0\
0 & b'+d'
end{bmatrix}=f(ax^3+bx^2+cx+d)+f(a'x^3+b'x^2+c'x+d')$



Is this correct?






linear-transformations







share|cite|improve this question













Determine if its linear the transformation



$f:Re_{3} [x]rightarrow M_{2x2}$



such as



$f(ax^3+bx^2+cx+d) = begin{bmatrix}
a-c & 0 \
0&b+d
end{bmatrix}$



for any $ax^3+bx^2+cx+d in Re _{3} [x]$



Im having trouble associating the polynomial with the resulting matrix.
I know I'm supposed to check the two main Linear transformation conditions but I don't know how to aproach this example.



Here's my attempt:



Condition 1: $T(u+v)=T(u)+T(v)$



$f((ax^3+bx^2+cx+d)+(a'x^3+b'x^2+c'x+d'))=f((ax^3+a'x^3)+(bx^2+b'x^2)+(cx+c'x)+(d+d'))=f((a+a')x^3+(b+b')x^2+(c+c')x+(d+d'))=begin{bmatrix}
(a+a')-(c+c') & 0\
0 & (b+b')+(d+d')
end{bmatrix} = begin{bmatrix}
a-c & 0\
0 & b+d
end{bmatrix} + begin{bmatrix}
a'-c' & 0\
0 & b'+d'
end{bmatrix}=f(ax^3+bx^2+cx+d)+f(a'x^3+b'x^2+c'x+d')$



Is this correct?






linear-transformations




linear-transformations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









JakcjonesJakcjones

588




588








  • 3




    Looks right to me.
    – saulspatz
    2 days ago






  • 3




    Seems fine. ${}{}{}{} $
    – Thomas Shelby
    2 days ago














  • 3




    Looks right to me.
    – saulspatz
    2 days ago






  • 3




    Seems fine. ${}{}{}{} $
    – Thomas Shelby
    2 days ago








3




3




Looks right to me.
– saulspatz
2 days ago




Looks right to me.
– saulspatz
2 days ago




3




3




Seems fine. ${}{}{}{} $
– Thomas Shelby
2 days ago




Seems fine. ${}{}{}{} $
– Thomas Shelby
2 days ago










1 Answer
1






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oldest

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2














That looks fine. Another way to go is this:



Consider the standard basis $$bigl{A_{1,1},A_{2,1},A_{1,2},A_{2,2}bigr}$$ for $M_{2times 2},$ where $A_{i,j}$ is the $2times 2$ matrix of zeroes, except the $i$th row $j$th column entry, which is $1.$ Considering also the standard basis ${x^3,x^2,x,1}$ for $mathfrak{R}_3[x],$ then the given transformation has the matrix representation $$begin{bmatrix}1 & 0 & -1 & 0\0 & 0 & 0 & 0\0 & 0 & 0 & 0\0 & 1 & 0 & 1end{bmatrix}.$$



Since it has a matrix representation, then it's a linear transformation.






share|cite|improve this answer





















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    1 Answer
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    1 Answer
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    active

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    2














    That looks fine. Another way to go is this:



    Consider the standard basis $$bigl{A_{1,1},A_{2,1},A_{1,2},A_{2,2}bigr}$$ for $M_{2times 2},$ where $A_{i,j}$ is the $2times 2$ matrix of zeroes, except the $i$th row $j$th column entry, which is $1.$ Considering also the standard basis ${x^3,x^2,x,1}$ for $mathfrak{R}_3[x],$ then the given transformation has the matrix representation $$begin{bmatrix}1 & 0 & -1 & 0\0 & 0 & 0 & 0\0 & 0 & 0 & 0\0 & 1 & 0 & 1end{bmatrix}.$$



    Since it has a matrix representation, then it's a linear transformation.






    share|cite|improve this answer


























      2














      That looks fine. Another way to go is this:



      Consider the standard basis $$bigl{A_{1,1},A_{2,1},A_{1,2},A_{2,2}bigr}$$ for $M_{2times 2},$ where $A_{i,j}$ is the $2times 2$ matrix of zeroes, except the $i$th row $j$th column entry, which is $1.$ Considering also the standard basis ${x^3,x^2,x,1}$ for $mathfrak{R}_3[x],$ then the given transformation has the matrix representation $$begin{bmatrix}1 & 0 & -1 & 0\0 & 0 & 0 & 0\0 & 0 & 0 & 0\0 & 1 & 0 & 1end{bmatrix}.$$



      Since it has a matrix representation, then it's a linear transformation.






      share|cite|improve this answer
























        2












        2








        2






        That looks fine. Another way to go is this:



        Consider the standard basis $$bigl{A_{1,1},A_{2,1},A_{1,2},A_{2,2}bigr}$$ for $M_{2times 2},$ where $A_{i,j}$ is the $2times 2$ matrix of zeroes, except the $i$th row $j$th column entry, which is $1.$ Considering also the standard basis ${x^3,x^2,x,1}$ for $mathfrak{R}_3[x],$ then the given transformation has the matrix representation $$begin{bmatrix}1 & 0 & -1 & 0\0 & 0 & 0 & 0\0 & 0 & 0 & 0\0 & 1 & 0 & 1end{bmatrix}.$$



        Since it has a matrix representation, then it's a linear transformation.






        share|cite|improve this answer












        That looks fine. Another way to go is this:



        Consider the standard basis $$bigl{A_{1,1},A_{2,1},A_{1,2},A_{2,2}bigr}$$ for $M_{2times 2},$ where $A_{i,j}$ is the $2times 2$ matrix of zeroes, except the $i$th row $j$th column entry, which is $1.$ Considering also the standard basis ${x^3,x^2,x,1}$ for $mathfrak{R}_3[x],$ then the given transformation has the matrix representation $$begin{bmatrix}1 & 0 & -1 & 0\0 & 0 & 0 & 0\0 & 0 & 0 & 0\0 & 1 & 0 & 1end{bmatrix}.$$



        Since it has a matrix representation, then it's a linear transformation.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Cameron BuieCameron Buie

        85.1k771155




        85.1k771155






























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