How to show that the Wronskian is independent to the choice of a basis of $V$? Let $f_1(x), \ldots, f_n(x)$ be a basis of a vector space $V$. Then the Wronskian is the determinant of a matrix whose i-th row is $f_1^{(i-1)}, \ldots, f_n^{(i-1)}$. Here $f_j^{(i)}$ is the i-th derivative of $f_j$. Thank you very much.
Wronskian is independent to the choice of a basis
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0What do you mean by $f_1(x),\ldots,f_n(x)$ is a basis of $V$? And in which space is $x$? – 2011-09-24
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1The statement in the title is false. If $n=1$ and $V$ is the space of all multiples of $f_1(x)$ then any $\lambda f_1(x)$, $\lambda\ne0$, is a basis of $V$, but the corresponding Wronskians $W_\lambda(x)=\lambda f_1(x)$ do not agree. – 2011-09-24
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1Maybe OP meant the orthonormal basis? – 2011-09-24
1 Answers
As stated in the comments, the Wronskian is not independent of basis, but it is determined up to a constant.
If $g_1(x),\dots,g_n(x)$ is a basis, then $g_i(x) = \sum_{j=1}^n c_{ij}f_j(x)$ for some constants $c_{ij}$ (since $f$'s form a basis). Let ${\bf f} = [f_1(x)\;f_2(x)\; \cdots \;f_n(x)]^T$, ${\bf g} = [g_1(x)\;g_2(x)\; \cdots \;g_n(x)]^T$, and $C = (c_{ij})_{n \times n}$. Then ${\bf g} = C{\bf f}$. Therefore, ${\bf g'} = C{\bf f'}$ (because $C$ is constant) and in general ${\bf g}^{(k)} = C{\bf f}^{(k)}$. Thus $[{\bf g} \; {\bf g}' \; \cdots \; {\bf g}^{(n-1)}] = [C{\bf f} \; C{\bf f}' \; \cdots \; C{\bf f}^{(n-1)}] = C[{\bf f} \; {\bf f}' \; \cdots \; {\bf f}^{(n-1)}]$. Therefore the Wronskian of the $g$'s is $\mathrm{det}(C)$ times the Wronskian of the $f$'s. [Also, since the $g$'s form a basis this determinant is a non-zero constant.]