coordinate vector relative to basis for $P_2$











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This was a question in one of my online assignments that I got wrong but have no idea why. Unfortunately I do not have the option to re-enter other attempts to see if they are correct, hence stack exchange.



Question:



Let p $=2x^2+6x+7$ .Find the coordinate vector of p relative to the following basis for $P_2$



$p_1 = 1$



$p_2 = 1+x$



$p_3 = 1+x+x^2$



My attempt:



Plugging in for simplicity $x=1$ for p , $p_1$, $p_2$, and $p_3$ results in the following vectors.



p = $(2,6,7)$



$p_1= $ $(1,0,0)$



$p_2= $ $(1,1,0)$



$p_3= $ $(1,1,1)$



Transposed of $p_n$ have:



$begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
*$begin{bmatrix}7\6\2end{bmatrix} = $$begin{bmatrix}15\8\2end{bmatrix}$



I reversed p because it was reversed relative to $p_n$






linear-algebra







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    up vote
    0
    down vote

    favorite












    This was a question in one of my online assignments that I got wrong but have no idea why. Unfortunately I do not have the option to re-enter other attempts to see if they are correct, hence stack exchange.



    Question:



    Let p $=2x^2+6x+7$ .Find the coordinate vector of p relative to the following basis for $P_2$



    $p_1 = 1$



    $p_2 = 1+x$



    $p_3 = 1+x+x^2$



    My attempt:



    Plugging in for simplicity $x=1$ for p , $p_1$, $p_2$, and $p_3$ results in the following vectors.



    p = $(2,6,7)$



    $p_1= $ $(1,0,0)$



    $p_2= $ $(1,1,0)$



    $p_3= $ $(1,1,1)$



    Transposed of $p_n$ have:



    $begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
    *$begin{bmatrix}7\6\2end{bmatrix} = $$begin{bmatrix}15\8\2end{bmatrix}$



    I reversed p because it was reversed relative to $p_n$






    linear-algebra







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      This was a question in one of my online assignments that I got wrong but have no idea why. Unfortunately I do not have the option to re-enter other attempts to see if they are correct, hence stack exchange.



      Question:



      Let p $=2x^2+6x+7$ .Find the coordinate vector of p relative to the following basis for $P_2$



      $p_1 = 1$



      $p_2 = 1+x$



      $p_3 = 1+x+x^2$



      My attempt:



      Plugging in for simplicity $x=1$ for p , $p_1$, $p_2$, and $p_3$ results in the following vectors.



      p = $(2,6,7)$



      $p_1= $ $(1,0,0)$



      $p_2= $ $(1,1,0)$



      $p_3= $ $(1,1,1)$



      Transposed of $p_n$ have:



      $begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
      *$begin{bmatrix}7\6\2end{bmatrix} = $$begin{bmatrix}15\8\2end{bmatrix}$



      I reversed p because it was reversed relative to $p_n$






      linear-algebra







      share|cite|improve this question













      This was a question in one of my online assignments that I got wrong but have no idea why. Unfortunately I do not have the option to re-enter other attempts to see if they are correct, hence stack exchange.



      Question:



      Let p $=2x^2+6x+7$ .Find the coordinate vector of p relative to the following basis for $P_2$



      $p_1 = 1$



      $p_2 = 1+x$



      $p_3 = 1+x+x^2$



      My attempt:



      Plugging in for simplicity $x=1$ for p , $p_1$, $p_2$, and $p_3$ results in the following vectors.



      p = $(2,6,7)$



      $p_1= $ $(1,0,0)$



      $p_2= $ $(1,1,0)$



      $p_3= $ $(1,1,1)$



      Transposed of $p_n$ have:



      $begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
      *$begin{bmatrix}7\6\2end{bmatrix} = $$begin{bmatrix}15\8\2end{bmatrix}$



      I reversed p because it was reversed relative to $p_n$






      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 Nov 28 at 1:38









      Forextrader

      346




      346






















          1 Answer
          1






          active

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          up vote
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          down vote



          accepted










          This is incorrect because you have simply taken the image of your polynomial under the matrix whose columns are the basis vectors.



          Instead, you want to augment that same matrix with the vector you were given: $left (begin{array}{rrr|r}1&1&1&7\0&1&1&6\0&0&1&2end{array}right)$.



          Now you can read off the solution. For starters, $z=2$.



          You are looking for the linear combination of the basis vectors which will give you $(7,6,2)^t$.






          share|cite|improve this answer























          • Ah ah ah, I see now. Thank you!
            – Forextrader
            Nov 28 at 1:59










          • But I was correct in changing (2,6,7) to (7,6,2) right?
            – Forextrader
            Nov 28 at 2:00










          • Yes. As the standard basis is${1,x,x^2}$.
            – Chris Custer
            Nov 28 at 2:08










          • In other words, you need to multiply the vector by the inverse of this matrix.
            – amd
            Nov 28 at 3:01










          • Yes that would do it.
            – Chris Custer
            Nov 28 at 3:16











          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          This is incorrect because you have simply taken the image of your polynomial under the matrix whose columns are the basis vectors.



          Instead, you want to augment that same matrix with the vector you were given: $left (begin{array}{rrr|r}1&1&1&7\0&1&1&6\0&0&1&2end{array}right)$.



          Now you can read off the solution. For starters, $z=2$.



          You are looking for the linear combination of the basis vectors which will give you $(7,6,2)^t$.






          share|cite|improve this answer























          • Ah ah ah, I see now. Thank you!
            – Forextrader
            Nov 28 at 1:59










          • But I was correct in changing (2,6,7) to (7,6,2) right?
            – Forextrader
            Nov 28 at 2:00










          • Yes. As the standard basis is${1,x,x^2}$.
            – Chris Custer
            Nov 28 at 2:08










          • In other words, you need to multiply the vector by the inverse of this matrix.
            – amd
            Nov 28 at 3:01










          • Yes that would do it.
            – Chris Custer
            Nov 28 at 3:16















          up vote
          1
          down vote



          accepted










          This is incorrect because you have simply taken the image of your polynomial under the matrix whose columns are the basis vectors.



          Instead, you want to augment that same matrix with the vector you were given: $left (begin{array}{rrr|r}1&1&1&7\0&1&1&6\0&0&1&2end{array}right)$.



          Now you can read off the solution. For starters, $z=2$.



          You are looking for the linear combination of the basis vectors which will give you $(7,6,2)^t$.






          share|cite|improve this answer























          • Ah ah ah, I see now. Thank you!
            – Forextrader
            Nov 28 at 1:59










          • But I was correct in changing (2,6,7) to (7,6,2) right?
            – Forextrader
            Nov 28 at 2:00










          • Yes. As the standard basis is${1,x,x^2}$.
            – Chris Custer
            Nov 28 at 2:08










          • In other words, you need to multiply the vector by the inverse of this matrix.
            – amd
            Nov 28 at 3:01










          • Yes that would do it.
            – Chris Custer
            Nov 28 at 3:16













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          This is incorrect because you have simply taken the image of your polynomial under the matrix whose columns are the basis vectors.



          Instead, you want to augment that same matrix with the vector you were given: $left (begin{array}{rrr|r}1&1&1&7\0&1&1&6\0&0&1&2end{array}right)$.



          Now you can read off the solution. For starters, $z=2$.



          You are looking for the linear combination of the basis vectors which will give you $(7,6,2)^t$.






          share|cite|improve this answer














          This is incorrect because you have simply taken the image of your polynomial under the matrix whose columns are the basis vectors.



          Instead, you want to augment that same matrix with the vector you were given: $left (begin{array}{rrr|r}1&1&1&7\0&1&1&6\0&0&1&2end{array}right)$.



          Now you can read off the solution. For starters, $z=2$.



          You are looking for the linear combination of the basis vectors which will give you $(7,6,2)^t$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 at 2:00

























          answered Nov 28 at 1:58









          Chris Custer

          10.2k3724




          10.2k3724












          • Ah ah ah, I see now. Thank you!
            – Forextrader
            Nov 28 at 1:59










          • But I was correct in changing (2,6,7) to (7,6,2) right?
            – Forextrader
            Nov 28 at 2:00










          • Yes. As the standard basis is${1,x,x^2}$.
            – Chris Custer
            Nov 28 at 2:08










          • In other words, you need to multiply the vector by the inverse of this matrix.
            – amd
            Nov 28 at 3:01










          • Yes that would do it.
            – Chris Custer
            Nov 28 at 3:16


















          • Ah ah ah, I see now. Thank you!
            – Forextrader
            Nov 28 at 1:59










          • But I was correct in changing (2,6,7) to (7,6,2) right?
            – Forextrader
            Nov 28 at 2:00










          • Yes. As the standard basis is${1,x,x^2}$.
            – Chris Custer
            Nov 28 at 2:08










          • In other words, you need to multiply the vector by the inverse of this matrix.
            – amd
            Nov 28 at 3:01










          • Yes that would do it.
            – Chris Custer
            Nov 28 at 3:16
















          Ah ah ah, I see now. Thank you!
          – Forextrader
          Nov 28 at 1:59




          Ah ah ah, I see now. Thank you!
          – Forextrader
          Nov 28 at 1:59












          But I was correct in changing (2,6,7) to (7,6,2) right?
          – Forextrader
          Nov 28 at 2:00




          But I was correct in changing (2,6,7) to (7,6,2) right?
          – Forextrader
          Nov 28 at 2:00












          Yes. As the standard basis is${1,x,x^2}$.
          – Chris Custer
          Nov 28 at 2:08




          Yes. As the standard basis is${1,x,x^2}$.
          – Chris Custer
          Nov 28 at 2:08












          In other words, you need to multiply the vector by the inverse of this matrix.
          – amd
          Nov 28 at 3:01




          In other words, you need to multiply the vector by the inverse of this matrix.
          – amd
          Nov 28 at 3:01












          Yes that would do it.
          – Chris Custer
          Nov 28 at 3:16




          Yes that would do it.
          – Chris Custer
          Nov 28 at 3:16


















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