Answer to Question #325920 in Linear Algebra for Lucifer

Question

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

Answer 1

Suppose that there are complex numbers $\alpha_1,\alpha_2,\alpha_3$ that are not all equal to zero and satisfy the equality: $\alpha_1(1+x)+\alpha_2(1+x^2)+\alpha_3(1+x^3)=0$ for all $x$. 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 $x=0,x=2,x=3.$ We receive equations $\alpha_1+\alpha_2+\alpha_3=0$, $3\alpha_1+5\alpha_2+9\alpha_3=0$, $4\alpha_1+10\alpha_2+28\alpha_3=0$. Since $\left|\begin{array}{lll}1&1&1\\3&5&9\\4&10&28\end{array}\right|=28\neq0$, the latter system of equations has the unique solution $\alpha_1=\alpha_2=\alpha_3=0$. Thus, due to the deduced contradiction, vectors $1+x,1+x^2,1+x^3$ are linearly independent.