Question #325920

Check whether the set of vectors {1 + 𝑥, 𝑥 + 𝑥2


, 1 + 𝑥3


} is a linearly independent set of


vectors in P3


, the vector space of polynomials of degree ≤ 3

1
Expert's answer
2022-04-11T07:36:57-0400

Suppose that there are complex numbers α1,α2,α3\alpha_1,\alpha_2,\alpha_3 that are not all equal to zero and satisfy the equality: α1(1+x)+α2(1+x2)+α3(1+x3)=0\alpha_1(1+x)+\alpha_2(1+x^2)+\alpha_3(1+x^3)=0 for all xx. We came to contradiction because a cubic equation may have only finite number of roots. Another way to come to a contradiction is to substitute values x=0,x=2,x=3.x=0,x=2,x=3. We receive equations α1+α2+α3=0\alpha_1+\alpha_2+\alpha_3=0, 3α1+5α2+9α3=03\alpha_1+5\alpha_2+9\alpha_3=0, 4α1+10α2+28α3=04\alpha_1+10\alpha_2+28\alpha_3=0. Since 11135941028=280\left|\begin{array}{lll}1&1&1\\3&5&9\\4&10&28\end{array}\right|=28\neq0, the latter system of equations has the unique solution α1=α2=α3=0\alpha_1=\alpha_2=\alpha_3=0. Thus, due to the deduced contradiction, vectors 1+x,1+x2,1+x31+x,1+x^2,1+x^3 are linearly independent.


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