Author Topic: Transforming a system of linear equations to a single higher-order equation  (Read 68 times)

nadia.chigmaroff

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Hi,
In the textbook, it says that a system of first-order equations can sometimes be transformed into a single higher-order equation (by the process given in problem 7 on the 7.1 problems).
Am I correct in saying that this is only possible to do in general when the determinant of the coefficient matrix associated with the system is nonzero?
I.e. if $\left\{\begin{aligned}
&x'_1= ax_1 + bx_2\\
&x'_2= cx_1 + dx_2
\end{aligned}\right.$, this can be converted into a singular equation iff $ad - bc$ $\neq 0$?
Thank you!   :D

aremorov

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No, this is still possible to do if the determinant is 0.

For example:

$x_1' = x_1 + x_2$ (*)
$x_2' = x_1 + x_2$

has determinant 0 for the coefficients, however if we isolate for $x_1$ we get:
$x_1'' - 2x' = 0$ which has solution:
$x_1 = C_1 + C_2 e^{2t}$ and putting this into equation (*) gives us $x_2 = C_2 e^{2t} -C_1$ for arbitrary constants $C_1, C_2$.

Victor Ivrii

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Nothing to do with the determinant. In dimension 2 you can reduce if the matrix is not diagonal--obvious. What about the diagonal case? It is also possible unless a matrix is scalar, i.e. proportional to identity. To do this reduction, we make first a linear transform, so that after it the matrix is not  diagonal anymore, and then reduce.

General criteria: System with constant coefficients could be reduced to a single equation iff each eigenspace is $1$-dimensional. To understand why we need to consider solutions to a homogeneous equation and to a homogeneous system.

For an equation one of the solutions is $t^{m-1} e^{kt}$ where $k$ is characteristic root, and $m$ is a multiplicity of $k$.

For system all solutions are in the form  $P_{s-1}(t)e^{kt}$ where $P_{s-1}(t)$ are polynomials of degree $\le s-1$ with vector-coefficients and $s$ is the maximal dimension of the corresponding Jordan cells.

Therefore reduction can be done iff $s=m$ which means that for each eigenvalue $k$ there is just one cell, which in turn means, that there is only one linearly independent eigenvector.

Remark: If $s_1,...,s_j$ are dimensions of all cells, corresponding to $k$, then their sum $=m$ where $m$ is a multiplicity of $k$ as a root of characteristic equation, and also the dimension of the root subspace, and $j$ is a dimension of the corresponding eigenspace.




However, it is not important: we solve systems without reducing them to  single equations.
« Last Edit: November 06, 2019, 01:06:21 AM by Victor Ivrii »

nadia.chigmaroff

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Ok, thanks!