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I'm studying a course in dynamical systems. It's a pretty much linear algebra intensive course, and it's been a while since I did that sort of things.

In it, they say that if vector $z$ satisfies $Az=\lambda z$, then $x(t) = c e^{\lambda t} z$ is a solution to $dx/dt = Ax$.

So far so good.

Further, they claim that if there are several solutions to the equation $Az = \lambda z$, that is $A z_1 = \lambda_1 z_1, A z_2 = \lambda_2 z_2, ... , A z_p = \lambda_p z_p$, then according to the superposition principle it follows that for all $c_1, c_2, ... , c_p$ the vector function $x(t) = c_1*e^{\lambda_1 t}z_1+c_2*e^{\lambda_2 t} z_2+...+c_p*e^{\lambda_p t}z_p$ is a solution to $dx/dt=Ax$. I don't understand this reasoning. Could anybody explain why it follows from the superposition principle. Tried all day get my head around it.

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Also, I must say I'm a little bit unsure what x(t) is supposed to mean here. Is it a vector or a scalar? I'm almost 100% sure at least t is a scalar (because I guess it is time). If it is a vector, what is the meaning of $dx/dt$? Do you just take the derivative of x elementwise? –  user1661303 May 6 '13 at 16:26
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Yes, x(t) is a vector function: if $A$ is an $n\times n$ matrix, then $x(t)\in\mathbb R^n$. The rest is a consequence of linearity. –  Ted Shifrin May 6 '13 at 16:27
    
It's because the differential equation is linear in $x$. –  Milind May 6 '13 at 16:27
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You are correct about $dx/dt$ - it is element-wise. If $x(t)$ were the location of a particle at time $t$, then $dx/dt$ would be the velocity (speed and direction) of the particle at time $t$. –  Thomas Andrews May 6 '13 at 16:29
    
That's the thing, I guess I'm not 100% sure about the definition of the derivative of a vector. $c_1*e^{\lambda1*t} z_1$ should be a vector since $z_1$ is a vector. Is it just elementwise? –  user1661303 May 6 '13 at 16:30
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1 Answer

up vote 2 down vote accepted

Given your final expression for $x(t)$,

$\displaystyle\frac{dx}{dt}=c_1\lambda_1 e^{\lambda_1 t} z_1+ c_2\lambda_2 e^{\lambda_2 t} z_2+ ... + c_p\lambda_p e^{\lambda_p t} z_p$.

Given your immediately preceding expressions relating $A$, $z_1, \lambda_1$, $z_2, \lambda_2$ etc.,

$\displaystyle\frac{dx}{dt}=c_1 e^{\lambda_1 t} Az_1 + c_2 e^{\lambda_2 t} Az_2 + ... + c_p e^{\lambda_p t} Az_p = A(c_1 e^{\lambda_1 t} z_1 + c_2 e^{\lambda_2 t} z_2 + ... + c_p e^{\lambda_p t} z_p)=$

$=Ax$

So your superposed expression is a solution of $\displaystyle\frac{dx}{dt}=Ax$.

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Still not getting it. The prerequisites is that $c_1*e^{\lambda_1 t}z_1 = dx/dt, c_2*e^{\lambda_2 t}z_2 = dx/dt, c_3*e^{\lambda_3 t}z_3 = dx/dt$. Why does that mean that the sum of them equals $dx/dt$? –  user1661303 May 6 '13 at 16:58
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Assume $x(t)=c_1 e^{\lambda_1 t} z_1 + ....$ Now take the time derivative of this expression. Now you should be able to see how to replace all the $\lambda_n$'s in the time derivative with $A$. Now factor out the $A$ and you should be able to see that you have $Ax$. So $dx/dt = Ax$. –  bob.sacamento May 6 '13 at 17:11
    
Actually, that makes sence. Thanks. –  user1661303 May 6 '13 at 17:17
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Glad to be of help. An upvote or an accept would be nice. Sorry to self-promote, but it's not like I'm the only one who does it. :) –  bob.sacamento May 6 '13 at 18:48
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@user1661303 Please consider upvote and accept the answer if you think it helps: meta.math.stackexchange.com/questions/3286/… –  Shuhao Cao Aug 3 '13 at 23:41
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