Wednesday, 05 October 2016 17:24

Linearly independency for exponential functions

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This is a supplementary notes for MATH211 ODE and Linear Algebra.

In the lecture, we discover that a good Ansatz for \(n\)-th order linear homogeneous ODE with constant coefficient is exponential functions. Indeed, if the characteristic function has \(n\) distinct real roots \(r_1,\cdots,r_n\). Then,  \(\{e^{r_1t},\cdots,e^{r_nt}\}\) are all solutions of the equation. If they are linearly independent, then they form a basis of the solution space, and therefore the general solution of the equation reads

\[y(t)=C_1e^{r_1t}+\cdots C_ne^{r_nt}.\]

Let us try to check whether they are always linearly independent.

One way to prove linearly independency of a set of smooth functions is through Wronskian. In this case, we have

\[W[e^{r_1t},\cdots,e^{r_nt}](t)=\text{det}\begin{bmatrix}e^{r_1t}&e^{r_2t}&\cdots&e^{r_nt}\\r_1e^{r_1t}&r_2e^{r_2t}&\cdots&r_ne^{r_nt}\\ \vdots&\vdots&\ddots&\vdots\\ r_1^{n-1}e^{r_1t}&r_2^{n-1}e^{r_2t}&\cdots&r_n^{n-1}e^{r_nt}\end{bmatrix}\]

In particular, if we evaluate the Wronskian at \(t=0\), what we get is

\[W[e^{r_1t},\cdots,e^{r_nt}](0)=\text{det}\begin{bmatrix}1&1&\cdots&1\\r_1&r_2&\cdots&r_n\\ \vdots&\vdots&\ddots&\vdots\\ r_1^{n-1}&r_2^{n-1}&\cdots&r_n^{n-1}\end{bmatrix}=\prod_{1\leq i<j\leq n}(r_j-r_i)\neq0.\]

Here, the matrix is realized as Vandermonde matrix [Wikipedia Page], and its determinant is always nonzero. So, we conclude that these functions are linearly independent and hence form a basis of the solution space.

The argument is also true for complex roots. For instance, one can prove \(\{\cos(r_1t),\cdots,\cos(r_nt)\}\) are linearly independent using a similar argument.

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