Answer on Question #79871 – Math – Linear Algebra
Question
Let P3 be the inner product space of polynomials of degree at most 3 over R with respect to the inner product
hf,gi =
Z 1
f(x)g(x)dx.
Apply the Gram-Schmidt orthogonalisation process to find an orthonormal basis for the subspace of P3 generated by the vectors (8)
1-2x,2x+6x
2
,-3x
2 +4x
3
Solution
a 1 = 1 − 2 x , a 2 = 2 x + 6 x 2 , a 3 = − 3 x 2 + 4 x 3 . a_1 = 1 - 2x, a_2 = 2x + 6x^2, a_3 = -3x^2 + 4x^3. a 1 = 1 − 2 x , a 2 = 2 x + 6 x 2 , a 3 = − 3 x 2 + 4 x 3 .
Apply the Gram-Schmidt orthogonalisation process:
b 1 = a 1 = 1 − 2 x , b_1 = a_1 = 1 - 2x, b 1 = a 1 = 1 − 2 x , b 2 = a 2 − ⟨ a 2 , b 1 ⟩ ⟨ b 1 , b 1 ⟩ b 1 . b_2 = a_2 - \frac{\langle a_2, b_1 \rangle}{\langle b_1, b_1 \rangle} b_1. b 2 = a 2 − ⟨ b 1 , b 1 ⟩ ⟨ a 2 , b 1 ⟩ b 1 . ⟨ a 2 , b 1 ⟩ = ∫ 0 1 ( 2 x + 6 x 2 ) ( 1 − 2 x ) d x = ∫ 0 1 ( 2 x + 2 x 2 − 12 x 3 ) d x = = 2 ⋅ 1 2 + 2 ⋅ 1 3 − 12 ⋅ 1 4 = − 4 3 \begin{array}{l}
\langle a_2, b_1 \rangle = \int_0^1 (2x + 6x^2)(1 - 2x)dx = \int_0^1 (2x + 2x^2 - 12x^3)dx = \\
= 2 \cdot \frac{1}{2} + 2 \cdot \frac{1}{3} - 12 \cdot \frac{1}{4} = -\frac{4}{3} \\
\end{array} ⟨ a 2 , b 1 ⟩ = ∫ 0 1 ( 2 x + 6 x 2 ) ( 1 − 2 x ) d x = ∫ 0 1 ( 2 x + 2 x 2 − 12 x 3 ) d x = = 2 ⋅ 2 1 + 2 ⋅ 3 1 − 12 ⋅ 4 1 = − 3 4 ⟨ b 1 , b 1 ⟩ = ∫ 0 1 ( 1 − 2 x ) 2 d x = ∫ 0 1 ( 1 − 4 x + 4 x 2 ) d x = = 1 − 4 ⋅ 1 2 + 4 ⋅ 1 3 = 1 3 \begin{array}{l}
\langle b_1, b_1 \rangle = \int_0^1 (1 - 2x)^2 dx = \int_0^1 (1 - 4x + 4x^2)dx = \\
= 1 - 4 \cdot \frac{1}{2} + 4 \cdot \frac{1}{3} = \frac{1}{3} \\
\end{array} ⟨ b 1 , b 1 ⟩ = ∫ 0 1 ( 1 − 2 x ) 2 d x = ∫ 0 1 ( 1 − 4 x + 4 x 2 ) d x = = 1 − 4 ⋅ 2 1 + 4 ⋅ 3 1 = 3 1 b 2 = 2 x + 6 x 2 − − 4 / 3 1 / 3 ( 1 − 2 x ) = 4 − 6 x + 6 x 2 b_2 = 2x + 6x^2 - \frac{-4/3}{1/3}(1 - 2x) = 4 - 6x + 6x^2 b 2 = 2 x + 6 x 2 − 1/3 − 4/3 ( 1 − 2 x ) = 4 − 6 x + 6 x 2 b 3 = a 3 − ⟨ a 3 , b 1 ⟩ ⟨ b 1 , b 1 ⟩ b 1 − ⟨ a 3 , b 2 ⟩ ⟨ b 2 , b 2 ⟩ b 2 b_3 = a_3 - \frac{\langle a_3, b_1 \rangle}{\langle b_1, b_1 \rangle} b_1 - \frac{\langle a_3, b_2 \rangle}{\langle b_2, b_2 \rangle} b_2 b 3 = a 3 − ⟨ b 1 , b 1 ⟩ ⟨ a 3 , b 1 ⟩ b 1 − ⟨ b 2 , b 2 ⟩ ⟨ a 3 , b 2 ⟩ b 2 ⟨ a 3 , b 1 ⟩ = ∫ 0 1 ( − 3 x 2 + 4 x 3 ) ( 1 − 2 x ) d x = ∫ 0 1 ( − 3 x 2 + 10 x 3 − 8 x 4 ) d x = = − 3 ⋅ 1 3 + 10 ⋅ 1 4 − 8 ⋅ 1 5 = − 1 10 \begin{array}{l}
\langle a_3, b_1 \rangle = \int_0^1 (-3x^2 + 4x^3)(1 - 2x)dx = \int_0^1 (-3x^2 + 10x^3 - 8x^4)dx = \\
= -3 \cdot \frac{1}{3} + 10 \cdot \frac{1}{4} - 8 \cdot \frac{1}{5} = -\frac{1}{10} \\
\end{array} ⟨ a 3 , b 1 ⟩ = ∫ 0 1 ( − 3 x 2 + 4 x 3 ) ( 1 − 2 x ) d x = ∫ 0 1 ( − 3 x 2 + 10 x 3 − 8 x 4 ) d x = = − 3 ⋅ 3 1 + 10 ⋅ 4 1 − 8 ⋅ 5 1 = − 10 1 ⟨ b 2 , b 2 ⟩ = ∫ 0 1 ( 4 − 6 x + 6 x 2 ) 2 d x = ∫ 0 1 ( 16 − 48 x + 84 x 2 − 72 x 3 + 36 x 4 ) d x = = 16 − 48 ⋅ 1 2 + 84 ⋅ 1 3 − 72 ⋅ 1 4 + 36 ⋅ 1 5 = 46 5 \begin{array}{l}
\left\langle b _ {2}, b _ {2} \right\rangle = \int_ {0} ^ {1} \left(4 - 6 x + 6 x ^ {2}\right) ^ {2} d x = \int_ {0} ^ {1} \left(1 6 - 4 8 x + 8 4 x ^ {2} - 7 2 x ^ {3} + 3 6 x ^ {4}\right) d x = \\
= 1 6 - 4 8 \cdot \frac {1}{2} + 8 4 \cdot \frac {1}{3} - 7 2 \cdot \frac {1}{4} + 3 6 \cdot \frac {1}{5} = \frac {4 6}{5} \\
\end{array} ⟨ b 2 , b 2 ⟩ = ∫ 0 1 ( 4 − 6 x + 6 x 2 ) 2 d x = ∫ 0 1 ( 16 − 48 x + 84 x 2 − 72 x 3 + 36 x 4 ) d x = = 16 − 48 ⋅ 2 1 + 84 ⋅ 3 1 − 72 ⋅ 4 1 + 36 ⋅ 5 1 = 5 46 b 3 = − 3 x 2 + 4 x 3 − − 1 / 10 1 / 3 ( 1 − 2 x ) − 1 / 10 46 / 5 ( 4 − 6 x + 6 x 2 ) = 59 230 − 123 230 x − 141 46 x 2 + 4 x 3 . b _ {3} = - 3 x ^ {2} + 4 x ^ {3} - \frac {- 1 / 1 0}{1 / 3} (1 - 2 x) - \frac {1 / 1 0}{4 6 / 5} (4 - 6 x + 6 x ^ {2}) = \frac {5 9}{2 3 0} - \frac {1 2 3}{2 3 0} x - \frac {1 4 1}{4 6} x ^ {2} + 4 x ^ {3}. b 3 = − 3 x 2 + 4 x 3 − 1/3 − 1/10 ( 1 − 2 x ) − 46/5 1/10 ( 4 − 6 x + 6 x 2 ) = 230 59 − 230 123 x − 46 141 x 2 + 4 x 3 .
We need to normalize all these vectors:
⟨ b 3 , b 3 ⟩ = 1 23 0 2 ∫ 0 1 ( 59 − 123 x − 705 x 2 + 920 x 3 ) 2 d x = = 1 23 0 2 ∫ 0 1 ( 5 9 2 + 12 3 2 x 2 + 70 5 2 x 4 + 92 0 2 x 6 − 2 ⋅ 59 ⋅ 123 x − 2 ⋅ 59 ⋅ 705 x 2 + 2 ⋅ 59 ⋅ 920 x 3 + + 2 ⋅ 123 ⋅ 705 x 3 − 2 ⋅ 123 ⋅ 920 x 4 − 2 ⋅ 705 ⋅ 920 x 5 ) d x = 40457 14 ⋅ 23 0 2 \begin{array}{l}
\left\langle b _ {3}, b _ {3} \right\rangle = \frac {1}{2 3 0 ^ {2}} \int_ {0} ^ {1} (5 9 - 1 2 3 x - 7 0 5 x ^ {2} + 9 2 0 x ^ {3}) ^ {2} d x = \\
= \frac {1}{2 3 0 ^ {2}} \int_ {0} ^ {1} \left(5 9 ^ {2} + 1 2 3 ^ {2} x ^ {2} + 7 0 5 ^ {2} x ^ {4} + 9 2 0 ^ {2} x ^ {6} - 2 \cdot 5 9 \cdot 1 2 3 x - 2 \cdot 5 9 \cdot 7 0 5 x ^ {2} + 2 \cdot 5 9 \cdot 9 2 0 x ^ {3} + \right. \\
\left. + 2 \cdot 1 2 3 \cdot 7 0 5 x ^ {3} - 2 \cdot 1 2 3 \cdot 9 2 0 x ^ {4} - 2 \cdot 7 0 5 \cdot 9 2 0 x ^ {5}\right) d x = \frac {4 0 4 5 7}{1 4 \cdot 2 3 0 ^ {2}} \\
\end{array} ⟨ b 3 , b 3 ⟩ = 23 0 2 1 ∫ 0 1 ( 59 − 123 x − 705 x 2 + 920 x 3 ) 2 d x = = 23 0 2 1 ∫ 0 1 ( 5 9 2 + 12 3 2 x 2 + 70 5 2 x 4 + 92 0 2 x 6 − 2 ⋅ 59 ⋅ 123 x − 2 ⋅ 59 ⋅ 705 x 2 + 2 ⋅ 59 ⋅ 920 x 3 + + 2 ⋅ 123 ⋅ 705 x 3 − 2 ⋅ 123 ⋅ 920 x 4 − 2 ⋅ 705 ⋅ 920 x 5 ) d x = 14 ⋅ 23 0 2 40457 e 1 = b 1 ⟨ b 1 , b 1 ⟩ = 1 − 2 x 1 / 3 = 3 − 2 3 x e _ {1} = \frac {b _ {1}}{\sqrt {\left\langle b _ {1} , b _ {1} \right\rangle}} = \frac {1 - 2 x}{\sqrt {1 / 3}} = \sqrt {3} - 2 \sqrt {3} x e 1 = ⟨ b 1 , b 1 ⟩ b 1 = 1/3 1 − 2 x = 3 − 2 3 x e 2 = b 2 ⟨ b 2 , b 2 ⟩ = 4 − 6 x + 6 x 2 46 / 5 = 2 5 46 ( 2 − 3 x + 3 x 2 ) e _ {2} = \frac {b _ {2}}{\sqrt {\left\langle b _ {2} , b _ {2} \right\rangle}} = \frac {4 - 6 x + 6 x ^ {2}}{\sqrt {4 6 / 5}} = 2 \sqrt {\frac {5}{4 6}} (2 - 3 x + 3 x ^ {2}) e 2 = ⟨ b 2 , b 2 ⟩ b 2 = 46/5 4 − 6 x + 6 x 2 = 2 46 5 ( 2 − 3 x + 3 x 2 ) e 3 = b 3 ⟨ b 3 , b 3 ⟩ = 59 − 123 x − 705 x 2 + 920 x 3 40457 / 14 = 14 40457 ( 59 − 123 x − 705 x 2 + 920 x 3 ) . e _ {3} = \frac {b _ {3}}{\sqrt {\left\langle b _ {3} , b _ {3} \right\rangle}} = \frac {5 9 - 1 2 3 x - 7 0 5 x ^ {2} + 9 2 0 x ^ {3}}{\sqrt {4 0 4 5 7 / 1 4}} = \sqrt {\frac {1 4}{4 0 4 5 7}} \left(5 9 - 1 2 3 x - 7 0 5 x ^ {2} + 9 2 0 x ^ {3}\right). e 3 = ⟨ b 3 , b 3 ⟩ b 3 = 40457/14 59 − 123 x − 705 x 2 + 920 x 3 = 40457 14 ( 59 − 123 x − 705 x 2 + 920 x 3 ) .
Answer:
e 1 = b 1 ⟨ b 1 , b 1 ⟩ = 1 − 2 x 1 / 3 = 3 − 2 3 x e _ {1} = \frac {b _ {1}}{\sqrt {\left\langle b _ {1} , b _ {1} \right\rangle}} = \frac {1 - 2 x}{\sqrt {1 / 3}} = \sqrt {3} - 2 \sqrt {3} x e 1 = ⟨ b 1 , b 1 ⟩ b 1 = 1/3 1 − 2 x = 3 − 2 3 x e 2 = b 2 ⟨ b 2 , b 2 ⟩ = 4 − 6 x + 6 x 2 46 / 5 = 2 5 46 ( 2 − 3 x + 3 x 2 ) e _ {2} = \frac {b _ {2}}{\sqrt {\left\langle b _ {2} , b _ {2} \right\rangle}} = \frac {4 - 6 x + 6 x ^ {2}}{\sqrt {4 6 / 5}} = 2 \sqrt {\frac {5}{4 6}} (2 - 3 x + 3 x ^ {2}) e 2 = ⟨ b 2 , b 2 ⟩ b 2 = 46/5 4 − 6 x + 6 x 2 = 2 46 5 ( 2 − 3 x + 3 x 2 ) e 3 = b 3 ⟨ b 3 , b 3 ⟩ = 59 − 123 x − 705 x 2 + 920 x 3 40457 / 14 = 14 40457 ( 59 − 123 x − 705 x 2 + 920 x 3 ) . e _ {3} = \frac {b _ {3}}{\sqrt {\left\langle b _ {3} , b _ {3} \right\rangle}} = \frac {5 9 - 1 2 3 x - 7 0 5 x ^ {2} + 9 2 0 x ^ {3}}{\sqrt {4 0 4 5 7 / 1 4}} = \sqrt {\frac {1 4}{4 0 4 5 7}} \left(5 9 - 1 2 3 x - 7 0 5 x ^ {2} + 9 2 0 x ^ {3}\right). e 3 = ⟨ b 3 , b 3 ⟩ b 3 = 40457/14 59 − 123 x − 705 x 2 + 920 x 3 = 40457 14 ( 59 − 123 x − 705 x 2 + 920 x 3 ) .
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#340153 on Dec 2023
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