Consider the system of ODEs

$$\textbf{y}'=A\textbf{y}, \ \ \ \textbf{y}(0)=\textbf{y}_0 $$ where A is a matrix. I have been given the matrix


And I am asked to prove that this has a unique solution under the assumption that the solution space is three-dimensional.

I am in an introductory level linear algebra class, and I am at the point where I find a general solution and a specific solution given the IVP conditions, however, I am unable to see how I am supposed to prove that it has a unique solution.

For reference, I have already solved the system for some initial value condition, that was part of an earlier problem. I just don't see how I am supposed to generalise anything.


1 Answer 1


$\det(\lambda I-A)=(\lambda-2)^2(\lambda +4).$ A basis for the eigenspace for eigenvalue 2 is $$\{\begin{bmatrix}1\\1\\0\end{bmatrix},\begin{bmatrix}-2\\0\\1\end{bmatrix}\}.$$ A basis for the eigenspace for eigenvalue -4 is $$\{\begin{bmatrix}-1\\-3\\2\end{bmatrix}\}.$$ Let $$P=\begin{bmatrix}1&-2&-1\\1&0&-3\\0&1&2\end{bmatrix}.$$ Let $$\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}=P\begin{bmatrix}u_1\\u_2\\u_3\end{bmatrix}$$ Since $P$ is non-singular, there is a 1-1 correspondence between particular values of $\mathbf y$ and particular values of $\mathbf u$. Moreover, $\mathbf u'=P^{-1}AP\mathbf u$,so the equations $$\mathbf y'=A\mathbf y, \mathbf y=\begin{bmatrix}a\\b\\c\end{bmatrix} \text{ when }t=0$$ have a unique solution iff the equations $$\mathbf u'=P^{-1}AP\mathbf u, \mathbf u=P^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix} \text{ when }t=0$$ have a unique solution. Write $$P^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}\ell\\m\\n\end{bmatrix}$$Note that $P^{-1}AP= \text { diag}(2,2,-4)$. The equations for $\mathbf u$ have a unique solution,$viz.$ $$u_1=\ell e^{2x},u_2=m e^{2x},u_3=n e^{-4x},$$ so the equations for $\mathbf y$ also have a unique silution.


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