Why can I use the eigenvectors of a matrix to "decouple" each coordinate contribution?

Question

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?

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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.

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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!

Answer 1

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.

Answer 2

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

Answer 3

IMHO, the main explanation or main answer to the question is missing.

The system of linear equations presented is a sub-space within a space of equations.

Like in any type of space, this space of equations will be characterized by it own set of dimensions, represented by any set of vectors, as long as these vectors form base for said space.

We all know that the eigenvectors could perfectly be such set of vectors, as they always form base for the sub-space defined by the matrix.

So, all the equations of said sub space of equations could be expressed as a linear combination of the eigenvectors, which since they form base, they are linearly dependent among them.

I hope this answer explains clearly and completely what is so special about eigenvectors.