Answer on Question #61909 – Math – Linear Algebra
Question
Let T : P 1 → P 2 T: P1 \to P2 T : P 1 → P 2
T ( a + b x ) = b + a x + ( a − b ) x 2 . T(a + bx) = b + ax + (a - b)x^2. T ( a + b x ) = b + a x + ( a − b ) x 2 .
Check that T T T B 1 = { 1 , x } B1 = \{1, x\} B 1 = { 1 , x } B 2 = { x 2 , x 2 + x , x 2 + x + 1 } B2 = \{x^2, x^2 + x, x^2 + x + 1\} B 2 = { x 2 , x 2 + x , x 2 + x + 1 } T T T T T T { a + b x + c x 2 ∈ P 2 ∣ a + c = b } \{a + bx + cx^2 \in P2 \mid a + c = b\} { a + b x + c x 2 ∈ P 2 ∣ a + c = b }
Solution
Let T ( a + b x ) = b + a x + ( a − b ) x 2 T(a + bx) = b + ax + (a - b)x^2 T ( a + b x ) = b + a x + ( a − b ) x 2 T ( a c + b d x ) = b d + a c x + ( a c − b d ) x 2 T(ac + bdx) = bd + acx + (ac - bd)x^2 T ( a c + b d x ) = b d + a c x + ( a c − b d ) x 2
c T ( a ) + d ( T ( b x ) ) = c ( a x + a x 2 ) + d ( b − b x 2 ) = b d + a c x + ( a c − b d ) x 2 = T ( a c + b d x ) cT(a) + d(T(bx)) = c(ax + ax^2) + d(b - bx^2) = bd + acx + (ac - bd)x^2 = T(ac + bdx) c T ( a ) + d ( T ( b x )) = c ( a x + a x 2 ) + d ( b − b x 2 ) = b d + a c x + ( a c − b d ) x 2 = T ( a c + b d x )
So, T T T
If B 1 = { 1 , x } = { ( 1 0 ) , ( 0 1 ) } B_1 = \{1, x\} = \left\{\binom{1}{0}, \binom{0}{1}\right\} B 1 = { 1 , x } = { ( 0 1 ) , ( 1 0 ) } B 2 = { x 2 , x 2 + x , x 2 + x + 1 } = { ( 1 0 ) , ( 0 1 ) , ( 0 1 ) } B_2 = \{x^2, x^2 + x, x^2 + x + 1\} = \left\{\binom{1}{0}, \binom{0}{1}, \binom{0}{1}\right\} B 2 = { x 2 , x 2 + x , x 2 + x + 1 } = { ( 0 1 ) , ( 1 0 ) , ( 1 0 ) }
T ( ( 1 0 ) ) = T ( 1 + 0 ⋅ x ) = x + x 2 = 0 ⋅ x 2 + 1 ⋅ ( x 2 + x ) + 0 ⋅ ( x 2 + x + 1 ) = 1 ⋅ ( 0 1 ) = ( 0 1 ) ; T\left(\binom{1}{0}\right) = T(1 + 0 \cdot x) = x + x^2 = 0 \cdot x^2 + 1 \cdot (x^2 + x) + 0 \cdot (x^2 + x + 1) = 1 \cdot \binom{0}{1} = \binom{0}{1}; T ( ( 0 1 ) ) = T ( 1 + 0 ⋅ x ) = x + x 2 = 0 ⋅ x 2 + 1 ⋅ ( x 2 + x ) + 0 ⋅ ( x 2 + x + 1 ) = 1 ⋅ ( 1 0 ) = ( 1 0 ) ; T ( ( 0 1 ) ) = T ( 0 + 1 ⋅ x ) = 1 − x 2 = − x 2 − ( x 2 + x ) + ( x 2 + x + 1 ) = − ( 1 0 ) − ( 0 1 ) + ( 0 0 ) = ( − 1 1 ) . T\left(\binom{0}{1}\right) = T(0 + 1 \cdot x) = 1 - x^2 = -x^2 - (x^2 + x) + (x^2 + x + 1) = -\binom{1}{0} - \binom{0}{1} + \binom{0}{0} = \binom{-1}{1}. T ( ( 1 0 ) ) = T ( 0 + 1 ⋅ x ) = 1 − x 2 = − x 2 − ( x 2 + x ) + ( x 2 + x + 1 ) = − ( 0 1 ) − ( 1 0 ) + ( 0 0 ) = ( 1 − 1 ) .
The matrix of the transformation T T T A = ( 0 − 1 1 − 1 0 1 ) A = \begin{pmatrix} 0 & -1 \\ 1 & -1 \\ 0 & 1 \end{pmatrix} A = ⎝ ⎛ 0 1 0 − 1 − 1 1 ⎠ ⎞
The following system determines the kernel of T T T
( 0 − 1 1 − 1 0 1 ) ( a b ) = ( 0 0 0 ) → − b = 0 ; b = 0 ; a − b = 0 → ( a = 0 b = 0 ) . \begin{pmatrix} 0 & -1 \\ 1 & -1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \to -b = 0; \, b = 0; \, a - b = 0 \to \begin{pmatrix} a = 0 \\ b = 0 \end{pmatrix}. ⎝ ⎛ 0 1 0 − 1 − 1 1 ⎠ ⎞ ( a b ) = ⎝ ⎛ 0 0 0 ⎠ ⎞ → − b = 0 ; b = 0 ; a − b = 0 → ( a = 0 b = 0 ) .
Thus, the kernel is K e r = ( 0 0 ) Ker = \binom{0}{0} Ker = ( 0 0 ) 0 + 0 ⋅ x = 0 0 + 0 \cdot x = 0 0 + 0 ⋅ x = 0
The range is T ( α + β x ) = β + α x + ( α − β ) x 2 = a + b x + c x 2 T(\alpha + \beta x) = \beta + \alpha x + (\alpha - \beta)x^2 = a + bx + cx^2 T ( α + β x ) = β + αx + ( α − β ) x 2 = a + b x + c x 2
where a = β ; b = α ; c = α − β a = \beta; b = \alpha; c = \alpha - \beta a = β ; b = α ; c = α − β
Thus, a + c = β + α − β = α = b a + c = \beta + \alpha - \beta = \alpha = b a + c = β + α − β = α = b { a + b x + c x 2 ∈ P 2 ∣ a + c = b } \{a + bx + cx^2 \in P_2 \mid a + c = b\} { a + b x + c x 2 ∈ P 2 ∣ a + c = b }
Answer: ( 0 − 1 1 − 1 0 1 ) \begin{pmatrix} 0 & -1 \\ 1 & -1 \\ 0 & 1 \end{pmatrix} ⎝ ⎛ 0 1 0 − 1 − 1 1 ⎠ ⎞ 0 , { a + b x + c x 2 ∈ P 2 ∣ a + c = b } 0, \{a + bx + cx^2 \in P_2 \mid a + c = b\} 0 , { a + b x + c x 2 ∈ P 2 ∣ a + c = b }
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