Find T(v) using the standard matrix and the matrix relative to B and B'












0












$begingroup$


Find T(V) by using the standard matrix and the matrix relative to B and B'



$T: R^3 -> R^2, T(x,y,z) = (x-y + 0, 0 + y-z), v = (2,4,6)$
$B = {(1,1,1), (1,1,0), (0,1,1)}$
$B' = {(1,1),(2,1)}$



Standard Matrix Way



A = $begin{bmatrix}1&-1&0 \ 0&1&-1end{bmatrix}$



$T(v)$ = $Av$ = $begin{bmatrix}1&-1&0 \ 0&1&-1end{bmatrix} begin{bmatrix}2 \ 4 \ 6end{bmatrix}$

= $begin{bmatrix}-2 \ -2 end{bmatrix}$



Matrix relative to B and B' Matrix



$T(v_1) = T(1,1,1) = (1, -1, 0) + (0, 1, -1) = (1,0, -1)$
$T(v_2) = T(1,1,0) = (1, -1, 0) + (0, 1, 0) = (1, 0, 0)$
$T(v_3) = T(0,1,1) = (0, -1, 0) + (0, 1, -1) = (0, 0, -1)$
$T(v)$ = $begin{bmatrix}1&1&0 \ 0&0&0 \ -1&0&-1end{bmatrix}$



$[v]_B$ = $2(1,1,1) + 4(1,1,0) + 6(0,1,1)$ = $(2,2,2) + (4,4,0) + (0,6,6)$
$[v]_B$ = $(6,14,8)$



$[T(v)]_B$ = $begin{bmatrix}1&1&0 \ 0&0&0 \ -1&0&-1end{bmatrix} begin{bmatrix}6 \ 14 \ 8end{bmatrix}$ = $begin{bmatrix}20 \ 0 \-14end{bmatrix}$



Obviously this is not the right answer, both should come out to be (-2,-2). What am I doing wrong?






linear-algebra







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$endgroup$








  • 1




    $begingroup$
    Your question has remarkably little detail. You just state a matrix and five points. What is $T(V)$ supposed to be, and what do the "Standard way" and "Matrix relative..." refer to? It's entirely unclear what you are trying to do :-)
    $endgroup$
    – Wolfgang Bangerth
    Dec 4 '18 at 15:01










  • $begingroup$
    The question simply says Find T(v) by using (1) the standard matrix and (2) the matrix relative to B and B'. T(v) is the linear transformation of vector v by multiplying the standard matrix * vector.
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 15:12


















0












$begingroup$


Find T(V) by using the standard matrix and the matrix relative to B and B'



$T: R^3 -> R^2, T(x,y,z) = (x-y + 0, 0 + y-z), v = (2,4,6)$
$B = {(1,1,1), (1,1,0), (0,1,1)}$
$B' = {(1,1),(2,1)}$



Standard Matrix Way



A = $begin{bmatrix}1&-1&0 \ 0&1&-1end{bmatrix}$



$T(v)$ = $Av$ = $begin{bmatrix}1&-1&0 \ 0&1&-1end{bmatrix} begin{bmatrix}2 \ 4 \ 6end{bmatrix}$

= $begin{bmatrix}-2 \ -2 end{bmatrix}$



Matrix relative to B and B' Matrix



$T(v_1) = T(1,1,1) = (1, -1, 0) + (0, 1, -1) = (1,0, -1)$
$T(v_2) = T(1,1,0) = (1, -1, 0) + (0, 1, 0) = (1, 0, 0)$
$T(v_3) = T(0,1,1) = (0, -1, 0) + (0, 1, -1) = (0, 0, -1)$
$T(v)$ = $begin{bmatrix}1&1&0 \ 0&0&0 \ -1&0&-1end{bmatrix}$



$[v]_B$ = $2(1,1,1) + 4(1,1,0) + 6(0,1,1)$ = $(2,2,2) + (4,4,0) + (0,6,6)$
$[v]_B$ = $(6,14,8)$



$[T(v)]_B$ = $begin{bmatrix}1&1&0 \ 0&0&0 \ -1&0&-1end{bmatrix} begin{bmatrix}6 \ 14 \ 8end{bmatrix}$ = $begin{bmatrix}20 \ 0 \-14end{bmatrix}$



Obviously this is not the right answer, both should come out to be (-2,-2). What am I doing wrong?






linear-algebra







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your question has remarkably little detail. You just state a matrix and five points. What is $T(V)$ supposed to be, and what do the "Standard way" and "Matrix relative..." refer to? It's entirely unclear what you are trying to do :-)
    $endgroup$
    – Wolfgang Bangerth
    Dec 4 '18 at 15:01










  • $begingroup$
    The question simply says Find T(v) by using (1) the standard matrix and (2) the matrix relative to B and B'. T(v) is the linear transformation of vector v by multiplying the standard matrix * vector.
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 15:12
















0












0








0





$begingroup$


Find T(V) by using the standard matrix and the matrix relative to B and B'



$T: R^3 -> R^2, T(x,y,z) = (x-y + 0, 0 + y-z), v = (2,4,6)$
$B = {(1,1,1), (1,1,0), (0,1,1)}$
$B' = {(1,1),(2,1)}$



Standard Matrix Way



A = $begin{bmatrix}1&-1&0 \ 0&1&-1end{bmatrix}$



$T(v)$ = $Av$ = $begin{bmatrix}1&-1&0 \ 0&1&-1end{bmatrix} begin{bmatrix}2 \ 4 \ 6end{bmatrix}$

= $begin{bmatrix}-2 \ -2 end{bmatrix}$



Matrix relative to B and B' Matrix



$T(v_1) = T(1,1,1) = (1, -1, 0) + (0, 1, -1) = (1,0, -1)$
$T(v_2) = T(1,1,0) = (1, -1, 0) + (0, 1, 0) = (1, 0, 0)$
$T(v_3) = T(0,1,1) = (0, -1, 0) + (0, 1, -1) = (0, 0, -1)$
$T(v)$ = $begin{bmatrix}1&1&0 \ 0&0&0 \ -1&0&-1end{bmatrix}$



$[v]_B$ = $2(1,1,1) + 4(1,1,0) + 6(0,1,1)$ = $(2,2,2) + (4,4,0) + (0,6,6)$
$[v]_B$ = $(6,14,8)$



$[T(v)]_B$ = $begin{bmatrix}1&1&0 \ 0&0&0 \ -1&0&-1end{bmatrix} begin{bmatrix}6 \ 14 \ 8end{bmatrix}$ = $begin{bmatrix}20 \ 0 \-14end{bmatrix}$



Obviously this is not the right answer, both should come out to be (-2,-2). What am I doing wrong?






linear-algebra







share|cite|improve this question









$endgroup$




Find T(V) by using the standard matrix and the matrix relative to B and B'



$T: R^3 -> R^2, T(x,y,z) = (x-y + 0, 0 + y-z), v = (2,4,6)$
$B = {(1,1,1), (1,1,0), (0,1,1)}$
$B' = {(1,1),(2,1)}$



Standard Matrix Way



A = $begin{bmatrix}1&-1&0 \ 0&1&-1end{bmatrix}$



$T(v)$ = $Av$ = $begin{bmatrix}1&-1&0 \ 0&1&-1end{bmatrix} begin{bmatrix}2 \ 4 \ 6end{bmatrix}$

= $begin{bmatrix}-2 \ -2 end{bmatrix}$



Matrix relative to B and B' Matrix



$T(v_1) = T(1,1,1) = (1, -1, 0) + (0, 1, -1) = (1,0, -1)$
$T(v_2) = T(1,1,0) = (1, -1, 0) + (0, 1, 0) = (1, 0, 0)$
$T(v_3) = T(0,1,1) = (0, -1, 0) + (0, 1, -1) = (0, 0, -1)$
$T(v)$ = $begin{bmatrix}1&1&0 \ 0&0&0 \ -1&0&-1end{bmatrix}$



$[v]_B$ = $2(1,1,1) + 4(1,1,0) + 6(0,1,1)$ = $(2,2,2) + (4,4,0) + (0,6,6)$
$[v]_B$ = $(6,14,8)$



$[T(v)]_B$ = $begin{bmatrix}1&1&0 \ 0&0&0 \ -1&0&-1end{bmatrix} begin{bmatrix}6 \ 14 \ 8end{bmatrix}$ = $begin{bmatrix}20 \ 0 \-14end{bmatrix}$



Obviously this is not the right answer, both should come out to be (-2,-2). What am I doing wrong?






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 14:54









Evan KimEvan Kim

998




998








  • 1




    $begingroup$
    Your question has remarkably little detail. You just state a matrix and five points. What is $T(V)$ supposed to be, and what do the "Standard way" and "Matrix relative..." refer to? It's entirely unclear what you are trying to do :-)
    $endgroup$
    – Wolfgang Bangerth
    Dec 4 '18 at 15:01










  • $begingroup$
    The question simply says Find T(v) by using (1) the standard matrix and (2) the matrix relative to B and B'. T(v) is the linear transformation of vector v by multiplying the standard matrix * vector.
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 15:12
















  • 1




    $begingroup$
    Your question has remarkably little detail. You just state a matrix and five points. What is $T(V)$ supposed to be, and what do the "Standard way" and "Matrix relative..." refer to? It's entirely unclear what you are trying to do :-)
    $endgroup$
    – Wolfgang Bangerth
    Dec 4 '18 at 15:01










  • $begingroup$
    The question simply says Find T(v) by using (1) the standard matrix and (2) the matrix relative to B and B'. T(v) is the linear transformation of vector v by multiplying the standard matrix * vector.
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 15:12










1




1




$begingroup$
Your question has remarkably little detail. You just state a matrix and five points. What is $T(V)$ supposed to be, and what do the "Standard way" and "Matrix relative..." refer to? It's entirely unclear what you are trying to do :-)
$endgroup$
– Wolfgang Bangerth
Dec 4 '18 at 15:01




$begingroup$
Your question has remarkably little detail. You just state a matrix and five points. What is $T(V)$ supposed to be, and what do the "Standard way" and "Matrix relative..." refer to? It's entirely unclear what you are trying to do :-)
$endgroup$
– Wolfgang Bangerth
Dec 4 '18 at 15:01












$begingroup$
The question simply says Find T(v) by using (1) the standard matrix and (2) the matrix relative to B and B'. T(v) is the linear transformation of vector v by multiplying the standard matrix * vector.
$endgroup$
– Evan Kim
Dec 4 '18 at 15:12






$begingroup$
The question simply says Find T(v) by using (1) the standard matrix and (2) the matrix relative to B and B'. T(v) is the linear transformation of vector v by multiplying the standard matrix * vector.
$endgroup$
– Evan Kim
Dec 4 '18 at 15:12












1 Answer
1






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1












$begingroup$

You shouldn't get the same answer: you should get $begin{pmatrix}-2\-2end{pmatrix}$expressed in terms of $beta '$.



So $begin{cases}x+2y=-2\x+y=-2end{cases}$.
So $x=-2,y=0$, that is, $begin{pmatrix}-2\0end{pmatrix}$.



Note $[T]_B^{B'}$ should be $2×3$.



Since $T(1,1,1)=(0,0)=0(1,1)+0(2,1), T(1,1,0)=(0,1)=2(1,1)-1(2,1)$ and $T(0,1,1)=(-1,0)=1(1,1)-1(2,1)$, we get $[T]_B^{B'}=begin{pmatrix}0&2&1\0&-1&-1end{pmatrix}$.



So, $(2,4,6)=4(1,1,1) -2(1,1,0)+2(0,1,1)$. Hence $begin{pmatrix}2\4\6end{pmatrix}_B=begin{pmatrix}4\-2\2end{pmatrix}$.



Finally check that $[T]_B^{B'}[v]_B=begin {pmatrix}-2\0end{pmatrix}=begin{pmatrix}-2\2end{pmatrix}_{B'}$.






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    $begingroup$

    You shouldn't get the same answer: you should get $begin{pmatrix}-2\-2end{pmatrix}$expressed in terms of $beta '$.



    So $begin{cases}x+2y=-2\x+y=-2end{cases}$.
    So $x=-2,y=0$, that is, $begin{pmatrix}-2\0end{pmatrix}$.



    Note $[T]_B^{B'}$ should be $2×3$.



    Since $T(1,1,1)=(0,0)=0(1,1)+0(2,1), T(1,1,0)=(0,1)=2(1,1)-1(2,1)$ and $T(0,1,1)=(-1,0)=1(1,1)-1(2,1)$, we get $[T]_B^{B'}=begin{pmatrix}0&2&1\0&-1&-1end{pmatrix}$.



    So, $(2,4,6)=4(1,1,1) -2(1,1,0)+2(0,1,1)$. Hence $begin{pmatrix}2\4\6end{pmatrix}_B=begin{pmatrix}4\-2\2end{pmatrix}$.



    Finally check that $[T]_B^{B'}[v]_B=begin {pmatrix}-2\0end{pmatrix}=begin{pmatrix}-2\2end{pmatrix}_{B'}$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You shouldn't get the same answer: you should get $begin{pmatrix}-2\-2end{pmatrix}$expressed in terms of $beta '$.



      So $begin{cases}x+2y=-2\x+y=-2end{cases}$.
      So $x=-2,y=0$, that is, $begin{pmatrix}-2\0end{pmatrix}$.



      Note $[T]_B^{B'}$ should be $2×3$.



      Since $T(1,1,1)=(0,0)=0(1,1)+0(2,1), T(1,1,0)=(0,1)=2(1,1)-1(2,1)$ and $T(0,1,1)=(-1,0)=1(1,1)-1(2,1)$, we get $[T]_B^{B'}=begin{pmatrix}0&2&1\0&-1&-1end{pmatrix}$.



      So, $(2,4,6)=4(1,1,1) -2(1,1,0)+2(0,1,1)$. Hence $begin{pmatrix}2\4\6end{pmatrix}_B=begin{pmatrix}4\-2\2end{pmatrix}$.



      Finally check that $[T]_B^{B'}[v]_B=begin {pmatrix}-2\0end{pmatrix}=begin{pmatrix}-2\2end{pmatrix}_{B'}$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You shouldn't get the same answer: you should get $begin{pmatrix}-2\-2end{pmatrix}$expressed in terms of $beta '$.



        So $begin{cases}x+2y=-2\x+y=-2end{cases}$.
        So $x=-2,y=0$, that is, $begin{pmatrix}-2\0end{pmatrix}$.



        Note $[T]_B^{B'}$ should be $2×3$.



        Since $T(1,1,1)=(0,0)=0(1,1)+0(2,1), T(1,1,0)=(0,1)=2(1,1)-1(2,1)$ and $T(0,1,1)=(-1,0)=1(1,1)-1(2,1)$, we get $[T]_B^{B'}=begin{pmatrix}0&2&1\0&-1&-1end{pmatrix}$.



        So, $(2,4,6)=4(1,1,1) -2(1,1,0)+2(0,1,1)$. Hence $begin{pmatrix}2\4\6end{pmatrix}_B=begin{pmatrix}4\-2\2end{pmatrix}$.



        Finally check that $[T]_B^{B'}[v]_B=begin {pmatrix}-2\0end{pmatrix}=begin{pmatrix}-2\2end{pmatrix}_{B'}$.






        share|cite|improve this answer









        $endgroup$



        You shouldn't get the same answer: you should get $begin{pmatrix}-2\-2end{pmatrix}$expressed in terms of $beta '$.



        So $begin{cases}x+2y=-2\x+y=-2end{cases}$.
        So $x=-2,y=0$, that is, $begin{pmatrix}-2\0end{pmatrix}$.



        Note $[T]_B^{B'}$ should be $2×3$.



        Since $T(1,1,1)=(0,0)=0(1,1)+0(2,1), T(1,1,0)=(0,1)=2(1,1)-1(2,1)$ and $T(0,1,1)=(-1,0)=1(1,1)-1(2,1)$, we get $[T]_B^{B'}=begin{pmatrix}0&2&1\0&-1&-1end{pmatrix}$.



        So, $(2,4,6)=4(1,1,1) -2(1,1,0)+2(0,1,1)$. Hence $begin{pmatrix}2\4\6end{pmatrix}_B=begin{pmatrix}4\-2\2end{pmatrix}$.



        Finally check that $[T]_B^{B'}[v]_B=begin {pmatrix}-2\0end{pmatrix}=begin{pmatrix}-2\2end{pmatrix}_{B'}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 18:08









        Chris CusterChris Custer

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