Find the change-of-coordinate matrices












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Let F be the field with 4 elements, F={0,1,x,y}. Let $V=P_3(F)$ be the space of polynomials p(t) of degree ≤3. (We can't use x as the polynomial variable since it is already used as an element of the field F; so we use t instead.)



a) Find the Lagrange interpolation polynomials $p_i(t)∈V, i=0,1,2,3$ corresponding to



$c_0=0, c_1=1, c_2=x, c_3=y$.



b) Consider the two bases of V given by



β={$1,t,t^2,t^3$}$, γ={p_0(t),p_1(t),p_2(t),p_3(t)}$
Find the change-of-coordinate matrices



$Q=[IV]^γ_β, P=[IV]^β_γ$.



part a) i did$$begin{bmatrix}0frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)} +1frac{(x-0)(x-2)(x-3)}{(1-1)(1-2)(1-3)} +xfrac{(x-0)(x-1)(x-3)}{(2-1)(2-2)(2-3)}+yfrac{(x-0)(x-1)(x-2)}{(3-1)(3-2)(3-3)} end{bmatrix}.$$
i have no idea abot part b, anyone can help






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    Let F be the field with 4 elements, F={0,1,x,y}. Let $V=P_3(F)$ be the space of polynomials p(t) of degree ≤3. (We can't use x as the polynomial variable since it is already used as an element of the field F; so we use t instead.)



    a) Find the Lagrange interpolation polynomials $p_i(t)∈V, i=0,1,2,3$ corresponding to



    $c_0=0, c_1=1, c_2=x, c_3=y$.



    b) Consider the two bases of V given by



    β={$1,t,t^2,t^3$}$, γ={p_0(t),p_1(t),p_2(t),p_3(t)}$
    Find the change-of-coordinate matrices



    $Q=[IV]^γ_β, P=[IV]^β_γ$.



    part a) i did$$begin{bmatrix}0frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)} +1frac{(x-0)(x-2)(x-3)}{(1-1)(1-2)(1-3)} +xfrac{(x-0)(x-1)(x-3)}{(2-1)(2-2)(2-3)}+yfrac{(x-0)(x-1)(x-2)}{(3-1)(3-2)(3-3)} end{bmatrix}.$$
    i have no idea abot part b, anyone can help






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      Let F be the field with 4 elements, F={0,1,x,y}. Let $V=P_3(F)$ be the space of polynomials p(t) of degree ≤3. (We can't use x as the polynomial variable since it is already used as an element of the field F; so we use t instead.)



      a) Find the Lagrange interpolation polynomials $p_i(t)∈V, i=0,1,2,3$ corresponding to



      $c_0=0, c_1=1, c_2=x, c_3=y$.



      b) Consider the two bases of V given by



      β={$1,t,t^2,t^3$}$, γ={p_0(t),p_1(t),p_2(t),p_3(t)}$
      Find the change-of-coordinate matrices



      $Q=[IV]^γ_β, P=[IV]^β_γ$.



      part a) i did$$begin{bmatrix}0frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)} +1frac{(x-0)(x-2)(x-3)}{(1-1)(1-2)(1-3)} +xfrac{(x-0)(x-1)(x-3)}{(2-1)(2-2)(2-3)}+yfrac{(x-0)(x-1)(x-2)}{(3-1)(3-2)(3-3)} end{bmatrix}.$$
      i have no idea abot part b, anyone can help






      linear-algebra







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      Let F be the field with 4 elements, F={0,1,x,y}. Let $V=P_3(F)$ be the space of polynomials p(t) of degree ≤3. (We can't use x as the polynomial variable since it is already used as an element of the field F; so we use t instead.)



      a) Find the Lagrange interpolation polynomials $p_i(t)∈V, i=0,1,2,3$ corresponding to



      $c_0=0, c_1=1, c_2=x, c_3=y$.



      b) Consider the two bases of V given by



      β={$1,t,t^2,t^3$}$, γ={p_0(t),p_1(t),p_2(t),p_3(t)}$
      Find the change-of-coordinate matrices



      $Q=[IV]^γ_β, P=[IV]^β_γ$.



      part a) i did$$begin{bmatrix}0frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)} +1frac{(x-0)(x-2)(x-3)}{(1-1)(1-2)(1-3)} +xfrac{(x-0)(x-1)(x-3)}{(2-1)(2-2)(2-3)}+yfrac{(x-0)(x-1)(x-2)}{(3-1)(3-2)(3-3)} end{bmatrix}.$$
      i have no idea abot part b, anyone can help






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      asked Nov 22 '18 at 2:03









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          Note that as indicated by the problem, $t$ should be used for polynomial variable, not $x$ (since we already use $x$ for element in the field $F$).



          I believe the problems in part (a) asks for four polynomials $p_i$ which form a (Lagrange) basis of $V$. In particular, $p_i(t)=displaystyleprod_{j=1,jne i}^3 frac{t-c_j}{c_i-c_j}$. For example, $p_1(t)=dfrac{(t-0)(t-x)(t-y)}{(1-0)(1-x)(1-y)}$ since $c_1=1$. (If you don't believe that these four are indeed a basis of $V$, you can try to prove it yourself or you can read the proof at here). I don't think your answer for part (a) is correct.



          For (b), the transition matrix from $beta= (1,t,t^2,t^3)$ to $beta'=(p_0(x),p_1(x),p_2(x),p_3(x))$ is exactly the Vandermonde matrix
          $P_{beta to beta'}= begin{pmatrix} 1 & 0 & 0 & 0 \
          1 & 1 & 1 & 1 \
          1 & x & x^2 & x^3 \
          1 & y & y^2 & y^3
          end{pmatrix}$          . In particular if a polynomial $f(t)=a_0+a_1t+a_2t^2+a_3t^3 in V$
          and $P_{betato beta'} (a_0,a_1,a_2,a_3)^T=(b_0,b_1,b_2,b_3)$ then
          $f(t)=b_0p_0(x)+b_1p_1(x)+b_2p_2(x)+b_3p_3(x)$.



          In fact, $boxed{b_i=f(a_i) text{ for
          all } 0 le i le 3}$           (hint: what is $p_j(a_i)$ for each $0 le i,j le 3$),
          which is the main reason why the above statement is true.






          share|cite|improve this answer























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            Note that as indicated by the problem, $t$ should be used for polynomial variable, not $x$ (since we already use $x$ for element in the field $F$).



            I believe the problems in part (a) asks for four polynomials $p_i$ which form a (Lagrange) basis of $V$. In particular, $p_i(t)=displaystyleprod_{j=1,jne i}^3 frac{t-c_j}{c_i-c_j}$. For example, $p_1(t)=dfrac{(t-0)(t-x)(t-y)}{(1-0)(1-x)(1-y)}$ since $c_1=1$. (If you don't believe that these four are indeed a basis of $V$, you can try to prove it yourself or you can read the proof at here). I don't think your answer for part (a) is correct.



            For (b), the transition matrix from $beta= (1,t,t^2,t^3)$ to $beta'=(p_0(x),p_1(x),p_2(x),p_3(x))$ is exactly the Vandermonde matrix
            $P_{beta to beta'}= begin{pmatrix} 1 & 0 & 0 & 0 \
            1 & 1 & 1 & 1 \
            1 & x & x^2 & x^3 \
            1 & y & y^2 & y^3
            end{pmatrix}$            . In particular if a polynomial $f(t)=a_0+a_1t+a_2t^2+a_3t^3 in V$
            and $P_{betato beta'} (a_0,a_1,a_2,a_3)^T=(b_0,b_1,b_2,b_3)$ then
            $f(t)=b_0p_0(x)+b_1p_1(x)+b_2p_2(x)+b_3p_3(x)$.



            In fact, $boxed{b_i=f(a_i) text{ for
            all } 0 le i le 3}$             (hint: what is $p_j(a_i)$ for each $0 le i,j le 3$),
            which is the main reason why the above statement is true.






            share|cite|improve this answer




























              0














              Note that as indicated by the problem, $t$ should be used for polynomial variable, not $x$ (since we already use $x$ for element in the field $F$).



              I believe the problems in part (a) asks for four polynomials $p_i$ which form a (Lagrange) basis of $V$. In particular, $p_i(t)=displaystyleprod_{j=1,jne i}^3 frac{t-c_j}{c_i-c_j}$. For example, $p_1(t)=dfrac{(t-0)(t-x)(t-y)}{(1-0)(1-x)(1-y)}$ since $c_1=1$. (If you don't believe that these four are indeed a basis of $V$, you can try to prove it yourself or you can read the proof at here). I don't think your answer for part (a) is correct.



              For (b), the transition matrix from $beta= (1,t,t^2,t^3)$ to $beta'=(p_0(x),p_1(x),p_2(x),p_3(x))$ is exactly the Vandermonde matrix
              $P_{beta to beta'}= begin{pmatrix} 1 & 0 & 0 & 0 \
              1 & 1 & 1 & 1 \
              1 & x & x^2 & x^3 \
              1 & y & y^2 & y^3
              end{pmatrix}$              . In particular if a polynomial $f(t)=a_0+a_1t+a_2t^2+a_3t^3 in V$
              and $P_{betato beta'} (a_0,a_1,a_2,a_3)^T=(b_0,b_1,b_2,b_3)$ then
              $f(t)=b_0p_0(x)+b_1p_1(x)+b_2p_2(x)+b_3p_3(x)$.



              In fact, $boxed{b_i=f(a_i) text{ for
              all } 0 le i le 3}$               (hint: what is $p_j(a_i)$ for each $0 le i,j le 3$),
              which is the main reason why the above statement is true.






              share|cite|improve this answer


























                0












                0








                0






                Note that as indicated by the problem, $t$ should be used for polynomial variable, not $x$ (since we already use $x$ for element in the field $F$).



                I believe the problems in part (a) asks for four polynomials $p_i$ which form a (Lagrange) basis of $V$. In particular, $p_i(t)=displaystyleprod_{j=1,jne i}^3 frac{t-c_j}{c_i-c_j}$. For example, $p_1(t)=dfrac{(t-0)(t-x)(t-y)}{(1-0)(1-x)(1-y)}$ since $c_1=1$. (If you don't believe that these four are indeed a basis of $V$, you can try to prove it yourself or you can read the proof at here). I don't think your answer for part (a) is correct.



                For (b), the transition matrix from $beta= (1,t,t^2,t^3)$ to $beta'=(p_0(x),p_1(x),p_2(x),p_3(x))$ is exactly the Vandermonde matrix
                $P_{beta to beta'}= begin{pmatrix} 1 & 0 & 0 & 0 \
                1 & 1 & 1 & 1 \
                1 & x & x^2 & x^3 \
                1 & y & y^2 & y^3
                end{pmatrix}$                . In particular if a polynomial $f(t)=a_0+a_1t+a_2t^2+a_3t^3 in V$
                and $P_{betato beta'} (a_0,a_1,a_2,a_3)^T=(b_0,b_1,b_2,b_3)$ then
                $f(t)=b_0p_0(x)+b_1p_1(x)+b_2p_2(x)+b_3p_3(x)$.



                In fact, $boxed{b_i=f(a_i) text{ for
                all } 0 le i le 3}$                 (hint: what is $p_j(a_i)$ for each $0 le i,j le 3$),
                which is the main reason why the above statement is true.






                share|cite|improve this answer














                Note that as indicated by the problem, $t$ should be used for polynomial variable, not $x$ (since we already use $x$ for element in the field $F$).



                I believe the problems in part (a) asks for four polynomials $p_i$ which form a (Lagrange) basis of $V$. In particular, $p_i(t)=displaystyleprod_{j=1,jne i}^3 frac{t-c_j}{c_i-c_j}$. For example, $p_1(t)=dfrac{(t-0)(t-x)(t-y)}{(1-0)(1-x)(1-y)}$ since $c_1=1$. (If you don't believe that these four are indeed a basis of $V$, you can try to prove it yourself or you can read the proof at here). I don't think your answer for part (a) is correct.



                For (b), the transition matrix from $beta= (1,t,t^2,t^3)$ to $beta'=(p_0(x),p_1(x),p_2(x),p_3(x))$ is exactly the Vandermonde matrix
                $P_{beta to beta'}= begin{pmatrix} 1 & 0 & 0 & 0 \
                1 & 1 & 1 & 1 \
                1 & x & x^2 & x^3 \
                1 & y & y^2 & y^3
                end{pmatrix}$                . In particular if a polynomial $f(t)=a_0+a_1t+a_2t^2+a_3t^3 in V$
                and $P_{betato beta'} (a_0,a_1,a_2,a_3)^T=(b_0,b_1,b_2,b_3)$ then
                $f(t)=b_0p_0(x)+b_1p_1(x)+b_2p_2(x)+b_3p_3(x)$.



                In fact, $boxed{b_i=f(a_i) text{ for
                all } 0 le i le 3}$                 (hint: what is $p_j(a_i)$ for each $0 le i,j le 3$),
                which is the main reason why the above statement is true.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 23 '18 at 10:20

























                answered Nov 23 '18 at 7:05









                TenguTengu

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