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Suppose I have a system of $n$ 1st order linear differential equations with constant coefficients. Then I can write that system of equations as $$ \frac{du}{dt} = Au$$ where $A$ is a constant matrix (no entries depend on time).

What is the reason I can use eigenvectors to decouple the system into $n$ equations wich don't depend on each other to be solved?

For example: If I have the following system of equations: $$ \frac{dx}{dt} = y $$ $$\frac{dy}{dt} = x$$ Then I can rewrite this as $$\frac{du}{dt}=\left( \begin{array}{ccc} 0 & 1 \\ 1 & 0 \end{array} \right)u$$ where $u = \left( \begin{array}{ccc} x \\ y \\ \end{array} \right)$

The eigenvectors of A are $\lambda_1=(1 , -1)^T$ and $(1,1)^T$. So, I could create new equations that don't depend on the other ones to solve them by using this eigenvectors: $$\frac{d(x+y)}{dt} = x+y $$ and $$\frac{d(x-y)}{dt} = -(x-y) $$

Therefore, I decoupled the equations.

I should be clear on something: I'm not looking for a proof of wheter this is the case. I'm looking for some insight on why does this happen with (almost) any matrix.

Hope I made myself clear, thank you!

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There is a slight difference in the resolution in the case the matrix is not diagonalizable. – user63181 Mar 7 '13 at 21:32

Hint: Assume a solution if the form $\vec{x}=e^{\lambda t}\vec{v}$ and see what happens.

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When you have $n$ linearly independent eigenvectors, $v_1,v_2,\dots,v_n$ we can write an invertible matrix $$E=\left(\begin{matrix}v_1&v_2&\dots&v_n\end{matrix}\right)$$ and $E^{-1}AE$ is a diagonal matrix.

Now, if $w=E^{-1}u$, then $\frac{dw}{dt} = E^{-1}\frac{du}{dt}$.

So $$E^{-1}AEw = E^{-1}Au = E^{-1}\frac{du}{dt} = \frac{dw}{dt}$$

So, if we can find $w$ which satisfies this new equation, we can get $u=Ew$, and visa versa.

But $E^{-1}AE$ is a diagonal matrix, so if $$w(t) = \left(\begin{matrix}w_1(t)\\w_2(t)\\\dots\\w_n(t)\end{matrix}\right)$$ we see that $\frac{dw}{dt} = E^{-1}AEw$ just means that $$\lambda_i w_i(t) = w_i'(t)$$ where $\lambda_i$ is the eigenvalue corresponding to $v_i$. And we know how to solve the simple equation equation. So we can solve for each $w_i$ and thus can solve for $u$.

Alternatively, if $u$ has a convergent Taylor expansion:

$$u(t) =\sum_{k=0}^\infty \frac{u^{(k)}(0)}{k!}t^k$$

Then we can see that that in general, $Au=u'$ means $Au^{(k)} = A^k u$, and therefore:

$$u(t) = \sum_{k=0}^\infty \frac{A^k u(0)}{k!} t^k = \left(\sum_{k=0}^\infty (tA)^k\right)u(0)=\exp(tA)u(0)$$

where $\exp(B)=\sum_{k=0}^\infty \frac{B^k}{k!}$ in the standard matrix exponentiation.

Now, eigenvectors of $A$ are eigenvectors of $\exp(A)$, and $\exp(T^{-1}AT)=T^{-1}\exp(A)T$ for invertible $T$. And if $A$ is diagonal with $\lambda_i$ along the diagonals, then $\exp(tA)$ is diagonal with $e^{\lambda_i t}$ along the diagonals.

So the general answer, even without enough eigenvectors, is that we want to find $\exp(tA)$. In the case of enough eigenvectors, $B=tE^{-1}AE$ is diagonal, and so $\exp(tA)=E\exp(tB)E^{-1}$, and $\exp(tB)$ is easy to express.

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