Find $T^*(x,y)$ with the given inner product











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Let $T:mathbb{C}^2 to mathbb{C}^2$ be a inner product space, $T(x,y)=(2ix+y, -x-iy)$ and the inner product $<(x_1, x_2), (y_1, y_2)>=4x_1overline{y_1}+9x_2overline{y_2}$.



Find $T^*(x,y)$



I found out that the canonical basis $C=left {(1,0), (0,1) right }$ isn't orthonormal (but orthogonal) for the given inner product, so I orthonormalized it and got $C'= left {(1/2,0), (0,1/3) right }$.



Then I used the definition for adjoint transformations:





  • $<T^*(1/2, 0), (x,y)>=<(1/2,0), T(x,y)>$



    $<(1/2, 0), (2ix+y, -x-iy)>=-4ix+2y$




  • $<T^*(0, 1/3), (x,y)>=<(0, 1/3), T(x,y)>$



    $<(0, 1/3), (2ix+y, -x-iy)>=-3x-3iy$




So $T^*(x,y)=(-4ix+2y, -3x-3iy)$, but the correct answer is $T^*(x,y)=(-2ix-9/4y, 4/9x+iy)$.



What am I doing wrong?






linear-algebra linear-transformations inner-product-space adjoint-operators







share|cite|improve this question






















  • I would use $T^*(x,y) = langle T^*(x,y), (1/2,0) rangle (1/2,0) + langle T^*(x,y), (0,1/3) rangle (0, 1/3)$.
    – Daniel Schepler
    Nov 16 at 23:14










  • Sorry, I still don't get the correct answer. Could you please explain a little more?
    – parishilton
    Nov 16 at 23:26










  • In your original question, how are you getting from the calculations of $langle T^*(1/2, 0), (x,y) rangle$ and $langle T^*(0, 1/3), (x,y) rangle$ (which should be $-4i bar x + 2 bar y$ and $-3 bar x - 3i bar y$ by the way) to a value for $T^*(x,y)$? I don't see the connection at all. At most, you could read off what $T^*(1/2,0)$ and $T^*(0,1/3)$ would be, and then say $T^*(x,y) = 2x T^*(1/2,0) + 3y T^*(0,1/3)$.
    – Daniel Schepler
    Nov 16 at 23:30










  • Ok, I get what you're saying, but now I have $T^*(x,y)=2x(-4i, -3)+3y(1, -3i)=(-8ix+3y, -6x-9iy)$? What am I doing wrong?
    – parishilton
    Nov 17 at 0:04










  • If $langle T^*(1/2, 0), (x,y) rangle = -4i bar x + 2 bar y = 4 (-i) bar x + 9 (frac{2}{9}) bar y = langle (-i, frac{2}{9}), (x,y) rangle$ for all $x,yin mathbb{C}$ then that would mean you must have $T^*(1/2, 0) = (-i, frac{2}{9})$...
    – Daniel Schepler
    Nov 17 at 0:13















up vote
2
down vote

favorite












Let $T:mathbb{C}^2 to mathbb{C}^2$ be a inner product space, $T(x,y)=(2ix+y, -x-iy)$ and the inner product $<(x_1, x_2), (y_1, y_2)>=4x_1overline{y_1}+9x_2overline{y_2}$.



Find $T^*(x,y)$



I found out that the canonical basis $C=left {(1,0), (0,1) right }$ isn't orthonormal (but orthogonal) for the given inner product, so I orthonormalized it and got $C'= left {(1/2,0), (0,1/3) right }$.



Then I used the definition for adjoint transformations:





  • $<T^*(1/2, 0), (x,y)>=<(1/2,0), T(x,y)>$



    $<(1/2, 0), (2ix+y, -x-iy)>=-4ix+2y$




  • $<T^*(0, 1/3), (x,y)>=<(0, 1/3), T(x,y)>$



    $<(0, 1/3), (2ix+y, -x-iy)>=-3x-3iy$




So $T^*(x,y)=(-4ix+2y, -3x-3iy)$, but the correct answer is $T^*(x,y)=(-2ix-9/4y, 4/9x+iy)$.



What am I doing wrong?






linear-algebra linear-transformations inner-product-space adjoint-operators







share|cite|improve this question






















  • I would use $T^*(x,y) = langle T^*(x,y), (1/2,0) rangle (1/2,0) + langle T^*(x,y), (0,1/3) rangle (0, 1/3)$.
    – Daniel Schepler
    Nov 16 at 23:14










  • Sorry, I still don't get the correct answer. Could you please explain a little more?
    – parishilton
    Nov 16 at 23:26










  • In your original question, how are you getting from the calculations of $langle T^*(1/2, 0), (x,y) rangle$ and $langle T^*(0, 1/3), (x,y) rangle$ (which should be $-4i bar x + 2 bar y$ and $-3 bar x - 3i bar y$ by the way) to a value for $T^*(x,y)$? I don't see the connection at all. At most, you could read off what $T^*(1/2,0)$ and $T^*(0,1/3)$ would be, and then say $T^*(x,y) = 2x T^*(1/2,0) + 3y T^*(0,1/3)$.
    – Daniel Schepler
    Nov 16 at 23:30










  • Ok, I get what you're saying, but now I have $T^*(x,y)=2x(-4i, -3)+3y(1, -3i)=(-8ix+3y, -6x-9iy)$? What am I doing wrong?
    – parishilton
    Nov 17 at 0:04










  • If $langle T^*(1/2, 0), (x,y) rangle = -4i bar x + 2 bar y = 4 (-i) bar x + 9 (frac{2}{9}) bar y = langle (-i, frac{2}{9}), (x,y) rangle$ for all $x,yin mathbb{C}$ then that would mean you must have $T^*(1/2, 0) = (-i, frac{2}{9})$...
    – Daniel Schepler
    Nov 17 at 0:13













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $T:mathbb{C}^2 to mathbb{C}^2$ be a inner product space, $T(x,y)=(2ix+y, -x-iy)$ and the inner product $<(x_1, x_2), (y_1, y_2)>=4x_1overline{y_1}+9x_2overline{y_2}$.



Find $T^*(x,y)$



I found out that the canonical basis $C=left {(1,0), (0,1) right }$ isn't orthonormal (but orthogonal) for the given inner product, so I orthonormalized it and got $C'= left {(1/2,0), (0,1/3) right }$.



Then I used the definition for adjoint transformations:





  • $<T^*(1/2, 0), (x,y)>=<(1/2,0), T(x,y)>$



    $<(1/2, 0), (2ix+y, -x-iy)>=-4ix+2y$




  • $<T^*(0, 1/3), (x,y)>=<(0, 1/3), T(x,y)>$



    $<(0, 1/3), (2ix+y, -x-iy)>=-3x-3iy$




So $T^*(x,y)=(-4ix+2y, -3x-3iy)$, but the correct answer is $T^*(x,y)=(-2ix-9/4y, 4/9x+iy)$.



What am I doing wrong?






linear-algebra linear-transformations inner-product-space adjoint-operators







share|cite|improve this question













Let $T:mathbb{C}^2 to mathbb{C}^2$ be a inner product space, $T(x,y)=(2ix+y, -x-iy)$ and the inner product $<(x_1, x_2), (y_1, y_2)>=4x_1overline{y_1}+9x_2overline{y_2}$.



Find $T^*(x,y)$



I found out that the canonical basis $C=left {(1,0), (0,1) right }$ isn't orthonormal (but orthogonal) for the given inner product, so I orthonormalized it and got $C'= left {(1/2,0), (0,1/3) right }$.



Then I used the definition for adjoint transformations:





  • $<T^*(1/2, 0), (x,y)>=<(1/2,0), T(x,y)>$



    $<(1/2, 0), (2ix+y, -x-iy)>=-4ix+2y$




  • $<T^*(0, 1/3), (x,y)>=<(0, 1/3), T(x,y)>$



    $<(0, 1/3), (2ix+y, -x-iy)>=-3x-3iy$




So $T^*(x,y)=(-4ix+2y, -3x-3iy)$, but the correct answer is $T^*(x,y)=(-2ix-9/4y, 4/9x+iy)$.



What am I doing wrong?






linear-algebra linear-transformations inner-product-space adjoint-operators




linear-algebra linear-transformations inner-product-space adjoint-operators






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 16 at 23:02









parishilton

1389




1389












  • I would use $T^*(x,y) = langle T^*(x,y), (1/2,0) rangle (1/2,0) + langle T^*(x,y), (0,1/3) rangle (0, 1/3)$.
    – Daniel Schepler
    Nov 16 at 23:14










  • Sorry, I still don't get the correct answer. Could you please explain a little more?
    – parishilton
    Nov 16 at 23:26










  • In your original question, how are you getting from the calculations of $langle T^*(1/2, 0), (x,y) rangle$ and $langle T^*(0, 1/3), (x,y) rangle$ (which should be $-4i bar x + 2 bar y$ and $-3 bar x - 3i bar y$ by the way) to a value for $T^*(x,y)$? I don't see the connection at all. At most, you could read off what $T^*(1/2,0)$ and $T^*(0,1/3)$ would be, and then say $T^*(x,y) = 2x T^*(1/2,0) + 3y T^*(0,1/3)$.
    – Daniel Schepler
    Nov 16 at 23:30










  • Ok, I get what you're saying, but now I have $T^*(x,y)=2x(-4i, -3)+3y(1, -3i)=(-8ix+3y, -6x-9iy)$? What am I doing wrong?
    – parishilton
    Nov 17 at 0:04










  • If $langle T^*(1/2, 0), (x,y) rangle = -4i bar x + 2 bar y = 4 (-i) bar x + 9 (frac{2}{9}) bar y = langle (-i, frac{2}{9}), (x,y) rangle$ for all $x,yin mathbb{C}$ then that would mean you must have $T^*(1/2, 0) = (-i, frac{2}{9})$...
    – Daniel Schepler
    Nov 17 at 0:13


















  • I would use $T^*(x,y) = langle T^*(x,y), (1/2,0) rangle (1/2,0) + langle T^*(x,y), (0,1/3) rangle (0, 1/3)$.
    – Daniel Schepler
    Nov 16 at 23:14










  • Sorry, I still don't get the correct answer. Could you please explain a little more?
    – parishilton
    Nov 16 at 23:26










  • In your original question, how are you getting from the calculations of $langle T^*(1/2, 0), (x,y) rangle$ and $langle T^*(0, 1/3), (x,y) rangle$ (which should be $-4i bar x + 2 bar y$ and $-3 bar x - 3i bar y$ by the way) to a value for $T^*(x,y)$? I don't see the connection at all. At most, you could read off what $T^*(1/2,0)$ and $T^*(0,1/3)$ would be, and then say $T^*(x,y) = 2x T^*(1/2,0) + 3y T^*(0,1/3)$.
    – Daniel Schepler
    Nov 16 at 23:30










  • Ok, I get what you're saying, but now I have $T^*(x,y)=2x(-4i, -3)+3y(1, -3i)=(-8ix+3y, -6x-9iy)$? What am I doing wrong?
    – parishilton
    Nov 17 at 0:04










  • If $langle T^*(1/2, 0), (x,y) rangle = -4i bar x + 2 bar y = 4 (-i) bar x + 9 (frac{2}{9}) bar y = langle (-i, frac{2}{9}), (x,y) rangle$ for all $x,yin mathbb{C}$ then that would mean you must have $T^*(1/2, 0) = (-i, frac{2}{9})$...
    – Daniel Schepler
    Nov 17 at 0:13
















I would use $T^*(x,y) = langle T^*(x,y), (1/2,0) rangle (1/2,0) + langle T^*(x,y), (0,1/3) rangle (0, 1/3)$.
– Daniel Schepler
Nov 16 at 23:14




I would use $T^*(x,y) = langle T^*(x,y), (1/2,0) rangle (1/2,0) + langle T^*(x,y), (0,1/3) rangle (0, 1/3)$.
– Daniel Schepler
Nov 16 at 23:14












Sorry, I still don't get the correct answer. Could you please explain a little more?
– parishilton
Nov 16 at 23:26




Sorry, I still don't get the correct answer. Could you please explain a little more?
– parishilton
Nov 16 at 23:26












In your original question, how are you getting from the calculations of $langle T^*(1/2, 0), (x,y) rangle$ and $langle T^*(0, 1/3), (x,y) rangle$ (which should be $-4i bar x + 2 bar y$ and $-3 bar x - 3i bar y$ by the way) to a value for $T^*(x,y)$? I don't see the connection at all. At most, you could read off what $T^*(1/2,0)$ and $T^*(0,1/3)$ would be, and then say $T^*(x,y) = 2x T^*(1/2,0) + 3y T^*(0,1/3)$.
– Daniel Schepler
Nov 16 at 23:30




In your original question, how are you getting from the calculations of $langle T^*(1/2, 0), (x,y) rangle$ and $langle T^*(0, 1/3), (x,y) rangle$ (which should be $-4i bar x + 2 bar y$ and $-3 bar x - 3i bar y$ by the way) to a value for $T^*(x,y)$? I don't see the connection at all. At most, you could read off what $T^*(1/2,0)$ and $T^*(0,1/3)$ would be, and then say $T^*(x,y) = 2x T^*(1/2,0) + 3y T^*(0,1/3)$.
– Daniel Schepler
Nov 16 at 23:30












Ok, I get what you're saying, but now I have $T^*(x,y)=2x(-4i, -3)+3y(1, -3i)=(-8ix+3y, -6x-9iy)$? What am I doing wrong?
– parishilton
Nov 17 at 0:04




Ok, I get what you're saying, but now I have $T^*(x,y)=2x(-4i, -3)+3y(1, -3i)=(-8ix+3y, -6x-9iy)$? What am I doing wrong?
– parishilton
Nov 17 at 0:04












If $langle T^*(1/2, 0), (x,y) rangle = -4i bar x + 2 bar y = 4 (-i) bar x + 9 (frac{2}{9}) bar y = langle (-i, frac{2}{9}), (x,y) rangle$ for all $x,yin mathbb{C}$ then that would mean you must have $T^*(1/2, 0) = (-i, frac{2}{9})$...
– Daniel Schepler
Nov 17 at 0:13




If $langle T^*(1/2, 0), (x,y) rangle = -4i bar x + 2 bar y = 4 (-i) bar x + 9 (frac{2}{9}) bar y = langle (-i, frac{2}{9}), (x,y) rangle$ for all $x,yin mathbb{C}$ then that would mean you must have $T^*(1/2, 0) = (-i, frac{2}{9})$...
– Daniel Schepler
Nov 17 at 0:13















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