# Show the uniqueness of solution of IVP

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

$$\begin{bmatrix}3&-1&2\\3&-1&6\\-2&2&-2\end{bmatrix}$$

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.

$$\det(\lambda I-A)=(\lambda-2)^2(\lambda +4).$$ 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.