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.

 

[blue_raver22]

I would approach this problem by first defining what it means for a set of functions to be linearly independent. In this case, it means that no function in the set can be written as a linear combination of the other functions. In other words, the only solution to the equation c1f1(x) + c2f2(x) + ... + cnfn(x) = 0 is when all the coefficients, c1, c2, ..., cn, are equal to zero.

Now, to show that the set of functions 1, x, x^2, ..., x^(n-1) is linearly independent using the Wronskian, we can utilize the definition of the Wronskian as the determinant of a matrix. This matrix will have the given functions as its rows, and the variables x and its derivatives as its columns.

Since the Wronskian is not equal to zero for at least one value in the interval, this means that the determinant of the matrix is not equal to zero. And since the determinant of a matrix is only equal to zero if and only if the matrix is singular (i.e. has no unique solution), this shows that the set of functions is linearly independent.

In conclusion, by using the Wronskian to show that the determinant of the matrix formed by the given functions is not equal to zero, we can prove that the set of functions 1, x, x^2, ..., x^(n-1) is linearly independent. This is a valid and rigorous way to show the linear independence of a set of functions.