3
$\begingroup$

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!

$\endgroup$
1
  • 2
    $\begingroup$ There is a slight difference in the resolution in the case the matrix is not diagonalizable. $\endgroup$
    – user63181
    Mar 7, 2013 at 21:32

3 Answers 3

1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ which since they form base, they are linearly independent among them. $\endgroup$
    – GEN
    Feb 21, 2017 at 15:59
  • $\begingroup$ I think this leads to a good intuition, but the explanation of it seems a bit obscure. $\endgroup$
    – David K
    Feb 21, 2017 at 16:07
  • $\begingroup$ For eigenvectorsthe inner product of any given pair of said vectors is always zero == zero projections. Since eigenvectors have zero "contribution" between them, there cannot be any kind of cross-contribution of a given coordinate and two or more eigenvectors, so, that clearly explains the reasons why eigenvectors offer a tool to "decouple" contributions of a given coordinate. Quod Erat Demonstrandum Kind regards, GEN $\endgroup$
    – GEN
    Jun 27, 2017 at 12:35
  • $\begingroup$ In a sense this is what we do when we use Principal Components Analysis to validate if the variables we are using in an experiment are linearly independent or not, and if some of them are not, by finding the eigenvectors and eigenvalues of the system, we find variables that are linearly independent, and their relative importance in terms of how much each one explains the variability of the dependent factor. Kind regards, GEN $\endgroup$
    – GEN
    Jun 27, 2017 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.