# Why use decoupling diagonalization method to solve system of ODE's instead of just the eigen vectors?

In my Advanced ODE's course we are learning about diagonalization of a coupled system of ODE's and using this to find the solution to the system such that the original system is:

$x'(t) = Ax$

So we find the eigen values and corresponding eigen vectors of matrix A and create a matrix P such that $P=[v_1 , v_2, ..., v_n]$ where the $v_i$'s are the eigen vectors as columns of matrix $P$, corresponding to eigen value $\lambda_i$.

Then we find matrix $B$ such that $B = PAP^{-1}$ and then B is diagonalized, with the previously found eigen values of $A$ on the diagonal and easily shows the new uncoupled system for, $y'(t) = By$, by change of variables, that can then be easily solved.

This will then result in a solution matrix, we call it $E(t)$, such that: $y(t) = E(t)y(0)$ which can then be used to find the general solution to the original coupled system of ODE's, $x'(t) = Ax$ such that $x(t) = PE(t)P^{-1}x(0)$.

My problem with understanding this method is why go through all the extra work to find the solution to the original system of ODE's by finding matrix B and using it to first solve the system $y'(t) = By$ to then finally solve the system $x'(t)=Ax$ when you have already found the eigen vectors and eigen values of $A$, and can use then to write the general solution, for example as:

$x(t) = c_1e^{\lambda_1t}[v_1] + c_2e^{\lambda_2t}[v_2]$

The diagonalization method just seems like extra unnecessary steps that I don't understand why we would take them?

The equation $x'(t) = A x(t)$ has solutions $$x(t) = e^{t A}\, x(0) \, ,$$ where $e^{t A}$ is the matrix exponential of $t A$. If $A$ is diagonalizable, i.e. $A = P B P^{-1}$ where $B = \text{diag}(\beta_i)$ is a diagonal matrix, then the matrix exponential can be computed according to $$e^{t A} = P e^{t B} P^{-1}\, , \qquad\text{where}\qquad e^{t B} = \text{diag}(e^{\beta_i t}) = E(t) \, .$$ Otherwise, it may be more difficult to compute the matrix exponential $e^{t A}$.
• Here you wrote $A$ as being diagonalizable, $A=UDU^{-1}$, but in the method I am using we create a whole new matrix $B$ that is diagonalized BY original matrix $A$, which is not necessarily originally already a diagonal matrix. Is this still the same thing? – TunaBooties Aug 31 '17 at 14:11
• @JustynaTafoya If I understood well, the matrix $B$ is diagonal since $\dot{y} = B y$ is uncoupled. Thus, it is the same as the diagonal matrix which I previously denoted by $D$. I have updated the notation to be consistent with the OP. – Harry49 Aug 31 '17 at 15:35