a) T ( ( P ( x ) ) = P ( x − 1 ) T((P(x))=P(x-1) T (( P ( x )) = P ( x − 1 )
T ( 1 ) = 0 , T ( x ) = x − 1 , T(1)=0 , T(x)=x-1, T ( 1 ) = 0 , T ( x ) = x − 1 ,
T ( x ) T(x) T ( x ) 2 = x =x = x 2 − 1 -1 − 1
With respect to U
0 = a ( 1 ) + b ( x ) + c ( x 0= a(1)+b(x)+c(x 0 = a ( 1 ) + b ( x ) + c ( x 2 ) ) )
a = 0 , b = 0 , c = 0 , a=0\:,b=0,c=0 , a = 0 , b = 0 , c = 0 ,
[ 0 ] u = [ 0 , 0 , 0 ] T [0]_u =[0,0,0]^T [ 0 ] u = [ 0 , 0 , 0 ] T
x − 1 = a ( 1 ) + b ( x ) + c ( x 2 ) x-1=a(1)+b(x)+c(x^2) x − 1 = a ( 1 ) + b ( x ) + c ( x 2 )
a = − 1 , b = 1 , c = 0 a=-1,b=1,c=0 a = − 1 , b = 1 , c = 0
[ x − 1 ] u = [ − 1 , 1 , 0 ] T [x-1]_u=[-1,1,0]^T [ x − 1 ] u = [ − 1 , 1 , 0 ] T
x 2 − 1 = a ( 1 ) + b ( x ) + c ( x 2 ) x^2-1=a(1)+b(x)+c(x^2) x 2 − 1 = a ( 1 ) + b ( x ) + c ( x 2 )
a = − 1 , b = 0 , c = 1 a=-1,b=0,c=1 a = − 1 , b = 0 , c = 1
[ x 2 − 1 ] u = [ − 1 , 0 , 1 ] T [x^2-1]_u=[-1,0,1]^T [ x 2 − 1 ] u = [ − 1 , 0 , 1 ] T
Matrix U _U U → \to → u(T) = ( 0 − 1 − 1 0 1 0 0 0 1 ) =\begin{pmatrix}
0&-1 & -1\\
0&1 & 0\\
0&0&1
\end{pmatrix} = ⎝ ⎛ 0 0 0 − 1 1 0 − 1 0 1 ⎠ ⎞
Part B
Coordu ( P ( x ) ) = [ 0 1 0 ] (P(x))= \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} ( P ( x )) = ⎣ ⎡ 0 1 0 ⎦ ⎤
Coord u ( T ( P ( x ) ) ) = (T(P(x)))= ( T ( P ( x ))) =
Coordu ( P ( x − 1 ) ) = [ − 1 1 0 ] (P(x-1))=\begin{bmatrix}
-1 \\
1\\
0
\end{bmatrix} ( P ( x − 1 )) = ⎣ ⎡ − 1 1 0 ⎦ ⎤
( 0 − 1 − 1 0 1 0 0 0 1 ) \begin{pmatrix}
0&-1 & -1\\
0&1 & 0\\
0&0&1
\end{pmatrix} ⎝ ⎛ 0 0 0 − 1 1 0 − 1 0 1 ⎠ ⎞ ( 0 1 0 ) \begin{pmatrix}
0\\
1\\
0
\end{pmatrix} ⎝ ⎛ 0 1 0 ⎠ ⎞ =( − 1 1 0 ) \begin{pmatrix}
-1 \\
1 \\
0
\end{pmatrix} ⎝ ⎛ − 1 1 0 ⎠ ⎞
Verified
Part C
Expressing elements of B in terms of the elements of U
1 + x + x 2 = a ( 1 ) + b ( x ) + c ( x 2 ) 1+x+x^2=a(1)+b(x)+c(x^2) 1 + x + x 2 = a ( 1 ) + b ( x ) + c ( x 2 )
a = 1 , b = 1 , c = 1 a=1,b=1,c=1 a = 1 , b = 1 , c = 1
[ 1 + x + x 2 ] u = [ 1 , 1 , 1 ] T [1+x+x^2]_u=[1,1,1]^T [ 1 + x + x 2 ] u = [ 1 , 1 , 1 ] T
2 x + x 2 = a ( 1 ) + b ( x ) + c ( x 2 ) 2x+x^2=a(1)+b(x)+c(x^2) 2 x + x 2 = a ( 1 ) + b ( x ) + c ( x 2 )
a = 0 , b = 2 , c = 1 a=0,b=2,c=1 a = 0 , b = 2 , c = 1
[ 2 x + x 2 ] u = [ 0 , 2 , 1 ] T [2x+x^2]_u=[0,2,1]^T [ 2 x + x 2 ] u = [ 0 , 2 , 1 ] T
x + x 2 = a ( 1 ) + b ( x ) + c ( x 2 ) x+x^2=a(1)+b(x)+c(x^2) x + x 2 = a ( 1 ) + b ( x ) + c ( x 2 )
a = 0 , b = 1 , c = 1 a=0,b=1,c=1 a = 0 , b = 1 , c = 1
[ x + x 2 ] u = [ 0 , 1 , 1 ] T [x+x^2]_u=[0,1,1]^T [ x + x 2 ] u = [ 0 , 1 , 1 ] T
Change of base matrix P B → U = P_B\to_U= P B → U =
[ 1 0 0 1 2 1 1 1 1 ] \begin{bmatrix}
1&0 & 0\\
1&2 & 1\\
1&1&1
\end{bmatrix} ⎣ ⎡ 1 1 1 0 2 1 0 1 1 ⎦ ⎤
Part D
Expressing elements of Coordu ( P ( x ) ) (P(x)) ( P ( x )) in terms of basis B
x = a ( 1 + x + x 2 ) + b ( 2 x + x 2 ) x=a(1+x+x^2)+b(2x+x^2) x = a ( 1 + x + x 2 ) + b ( 2 x + x 2 )
+ c ( x + x 2 ) +c(x+x^2) + c ( x + x 2 )
x = a + ( a + 2 b + c ) x + ( a + b + c ) x 2 x=a+(a+2b+c)x+(a+b+c)x^2 x = a + ( a + 2 b + c ) x + ( a + b + c ) x 2
⟹ a = 0 , a + 2 b + c = 1 \implies a=0,a+2b+c=1 ⟹ a = 0 , a + 2 b + c = 1
⟹ 2 b + c = 1....... ( i ) \implies 2b+c=1.......(i) ⟹ 2 b + c = 1....... ( i )
a + b + c = 0 a+b+c=0 a + b + c = 0
⟹ b + c = 0...... ( i i ) \implies b+c=0......(ii) ⟹ b + c = 0...... ( ii )
Solving ( i ) (i) ( i ) and( i i ) (ii) ( ii )
b = 1 , c = − 1 b=1,c=-1 b = 1 , c = − 1
CoordB ( P ( x ) ) = ( 0 1 − 1 ) _B(P(x))=\begin{pmatrix}
0 \\
1\\
-1
\end{pmatrix} B ( P ( x )) = ⎝ ⎛ 0 1 − 1 ⎠ ⎞
Alternatively (method 2 )
Coordu ( P ( x ) ) = ( 0 1 0 ) (P(x))=\begin{pmatrix}
0 \\
1 \\
0
\end{pmatrix} ( P ( x )) = ⎝ ⎛ 0 1 0 ⎠ ⎞
Change of basis matrix P u → B = P_u\to_B= P u → B =
( M a t r i x P B → u ) − 1 (Matrix P_B\to_u)^{-1} ( M a t r i x P B → u ) − 1
∴ ( 1 0 0 1 2 1 1 1 1 ) \therefore \begin{pmatrix}
1&0 & 0 \\
1&2 & 1\\
1&1&1
\end{pmatrix} ∴ ⎝ ⎛ 1 1 1 0 2 1 0 1 1 ⎠ ⎞ − 1 ^-1 − 1 ( coordB ( P ( x ) ) _B(P(x)) B ( P ( x )) )
( 1 0 0 0 1 − 1 − 1 − 1 2 ) \begin{pmatrix}
1&0& 0\\
0&1&-1\\
-1&-1&2
\end{pmatrix} ⎝ ⎛ 1 0 − 1 0 1 − 1 0 − 1 2 ⎠ ⎞ ( 0 1 0 ) \begin{pmatrix}
0 \\
1 \\
0
\end{pmatrix} ⎝ ⎛ 0 1 0 ⎠ ⎞ =( 0 1 − 1 ) \begin{pmatrix}
0\\
1\\
-1
\end{pmatrix} ⎝ ⎛ 0 1 − 1 ⎠ ⎞
Comments