Determine a real number such that the vectors given are linearly independent

Question

Given vectors $f_1(x) = e^x, f_2(x) =xe^x, f_3(x) =x^ke^x$ in vector space $C$. Determine the values of real number $k$ such that the vectors are linearly independent.

Note that if we write in its linear combination, we have $C_1e^x + C_2xe^x + C_3x^ke^x = 0$

Here we need to show that $C_1 = C_2 = C_3 = 0$ is the only solution for the equation above, hence we conclude that the vectors are linearly independent. But how can I show that? Should I show that the determinant of its coefficient matrix is not zero? Please show your help

Answer 1

We can calculate the wronskian of this functions. The theorem says that if $W(e^x, xe^x, x^k e^x)\neq 0$ then the functions are linearly independent: $$W(e^x, xe^x, x^k e^x)\neq 0$$

$$ \begin{vmatrix}e^x& xe^x & x^k e^x\\ e^x& e^x(x+1) & e^x(x^k +kx^{k-1})\\ e^x& e^x(x+2) & e^x(x^k + 2kx^{k-1} +k(k-1)x^{k-2})\\ \end{vmatrix}\neq 0$$

Applying Gauss reduction $$ \begin{vmatrix}e^x& xe^x & x^k e^x\\ 0& 1 & e^x(kx^{k-1})\\ 0& 2 & e^x( 2kx^{k-1} +k(k-1)x^{k-2})\\ \end{vmatrix}\neq 0$$ $$ e^x(e^x( 2kx^{k-1} +k(k-1)x^{k-2})-2e^x(kx^{k-1}))\neq0$$ $$ e^{2x}( 2kx^{k-1} +k(k-1)x^{k-2}-2kx^{k-1})\neq0$$ $$ (k(k-1)x^{k-2})\neq 0$$ Finally we get $k\notin \{0, 1\}$.

Answer $k\in \mathbb{R}\setminus\{0, 1\}$

Answer 2

This is a straightforward calculation.

First note that we obviously must exclude $k=0, 1$.

Now canel the factor of $e^x$ from your equation. Then you have $c_1+c_2x+c_3x^k=0$, which must hold for all values of $x$.

Setting $x=0$ shows $c_1=0$. Thus $x(c_2+c_3x^{k-1})=0$.

Setting $x=1$ shows $c_2=-c_3$.

Setting $x=2$ shows $c_2(1-2^{k-1})=0$. Since $k\neq1$, this shows $c_2=0$.

What does this tell you about the possible values of $k$?