Answer on Question #81142 – Math – Linear Algebra
Question
Complete the set S = { x S = \{x S = { x
3 + x
2 + 1, x
2 + x + 1, x + 1 \} to get a basis of P3
Solution
Denote
a 1 = x 3 + x 2 + 1 , a 2 = x 2 + x + 1 , a 3 = x + 1. a_1 = x^3 + x^2 + 1, a_2 = x^2 + x + 1, a_3 = x + 1. a 1 = x 3 + x 2 + 1 , a 2 = x 2 + x + 1 , a 3 = x + 1.
Let's check the hypothesis that a 1 , a 2 , a 3 a_1, a_2, a_3 a 1 , a 2 , a 3 and a 4 = 1 a_4 = 1 a 4 = 1 form a basis in P 3 P_3 P 3 . We have to check whether c 1 a 1 + c 2 a 2 + c 3 a 3 + c 4 a 4 = 0 c_1a_1 + c_2a_2 + c_3a_3 + c_4a_4 = 0 c 1 a 1 + c 2 a 2 + c 3 a 3 + c 4 a 4 = 0
yields
c 1 = c 2 = c 3 = c 4 = 0. c_1 = c_2 = c_3 = c_4 = 0. c 1 = c 2 = c 3 = c 4 = 0.
We have
c 1 ( x 3 + x 2 + 1 ) + c 2 ( x 2 + x + 1 ) + c 3 ( x + 1 ) + c 4 ⋅ 1 = 0 c_1(x^3 + x^2 + 1) + c_2(x^2 + x + 1) + c_3(x + 1) + c_4 \cdot 1 = 0 c 1 ( x 3 + x 2 + 1 ) + c 2 ( x 2 + x + 1 ) + c 3 ( x + 1 ) + c 4 ⋅ 1 = 0
This means
c 1 x 3 + ( c 1 + c 2 ) x 2 + ( c 2 + c 3 ) x + ( c 1 + c 2 + c 3 + c 4 ) = 0. c_1x^3 + (c_1 + c_2)x^2 + (c_2 + c_3)x + (c_1 + c_2 + c_3 + c_4) = 0. c 1 x 3 + ( c 1 + c 2 ) x 2 + ( c 2 + c 3 ) x + ( c 1 + c 2 + c 3 + c 4 ) = 0.
Then
{ c 1 = 0 c 1 + c 2 = 0 c 2 + c 3 = 0 c 1 + c 2 + c 3 + c 4 = 0 \left\{ \begin{array}{l}
c_1 = 0 \\
c_1 + c_2 = 0 \\
c_2 + c_3 = 0 \\
c_1 + c_2 + c_3 + c_4 = 0
\end{array} \right. ⎩ ⎨ ⎧ c 1 = 0 c 1 + c 2 = 0 c 2 + c 3 = 0 c 1 + c 2 + c 3 + c 4 = 0
from which
c 1 = 0 , c 2 = − c 1 = 0 , c 3 = − c 2 = 0 , c 4 = − c 1 − c 2 − c 3 = 0. c_1 = 0, c_2 = -c_1 = 0, c_3 = -c_2 = 0, c_4 = -c_1 - c_2 - c_3 = 0. c 1 = 0 , c 2 = − c 1 = 0 , c 3 = − c 2 = 0 , c 4 = − c 1 − c 2 − c 3 = 0.
This means that a 1 , a 2 , a 3 , a 4 a_1, a_2, a_3, a_4 a 1 , a 2 , a 3 , a 4 really form a basis in P 3 P_3 P 3 .
Answer: 1 should be added, the basis of P 3 P_3 P 3 is { x 3 + x 2 + 1 , x 2 + x + 1 , x + 1 , 1 } \{x^3 + x^2 + 1, x^2 + x + 1, x + 1, 1\} { x 3 + x 2 + 1 , x 2 + x + 1 , x + 1 , 1 } .
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