MHB Linearly independence using Wronskian?

[Cadbury]

Hi, so I am given this problem:

Using the Wronskian, show that 1, x, x^2,..., x^(n-1) for n>1 are linearly independent.

The wronskian is not zero for at least one value in the interval so it is linearly independent, I just do not know how to show it properly.Thank you! :D

 

[Euge]

Hi Cadbury,

Linear independence of a set of functions $f_1(x),\ldots, f_n(x)$ means that for every $x$ (not just a particular $x$), $f_1(x),\ldots, f_n(x)$ is linearly independent. To show that $1,x,\ldots, x^{n-1}$ is linearly independent using the Wronskian, you must show that the Wronskian $W(1,x,\ldots, x^{n-1})$ is never zero. Prove that $W(1,x,\ldots, x^{n-1})$ is the determinant of an upper triangular $n\times n$ matrix whose $j$th diagonal entry is $(j-1)!$. Explain why this implies $W(1,x,\ldots, x^{n-1})$ is never zero.

 

[Cadbury]

Hi Cadbury,

Linear independence of a set of functions $f_1(x),\ldots, f_n(x)$ means that for every $x$ (not just a particular $x$), $f_1(x),\ldots, f_n(x)$ is linearly independent. To show that $1,x,\ldots, x^{n-1}$ is linearly independent using the Wronskian, you must show that the Wronskian $W(1,x,\ldots, x^{n-1})$ is never zero. Prove that $W(1,x,\ldots, x^{n-1})$ is the determinant of an upper triangular $n\times n$ matrix whose $j$th diagonal entry is $(j-1)!$. Explain why this implies $W(1,x,\ldots, x^{n-1})$ is never zero.

so, i can just show a matrix then find the determinant and then it is done? :) Thank you!

 

[Euge]

Well, either by computing the Wronskian (getting a value of $2! 3!\cdots (n-1)!$) or by noting that the Wronskian matrix is upper triangular with nonzero diagonal entries (hence invertible), you deduce that $W(1,x,\ldots x^{n-1}) \neq 0$ for all $x$, which shows that $1,x,\ldots, x^{n-1}$ is linearly independent.