Given an LTI system of the form:

$$ \frac{d}{dt}{x}(t) = {A} {x}(t) \ + \ {B} {u}(t) $$

where ${x}(t)$ is the state variable, ${u}(t)$ an input and ${A}$ and ${B}$ constants, how does one show, by superposition, that the system is linear? I've searched and searched and tried to do it myself and I am unable to do so. Seems more difficult that it seems. Most books seem to gloss over the fact.

  • $\begingroup$ Most people expand LTI when they encounter the acronym in system theory as linear time-invariant and a linear system is by definition one that satisfies the superposition property. So, in some sense, there is nothing to prove. $\endgroup$ Commented Dec 28, 2014 at 4:33
  • $\begingroup$ i am editing the math in all of this to use proper matrix/vector notation that you get with state-variable system description. $\endgroup$ Commented Dec 28, 2014 at 20:02
  • $\begingroup$ with an LTI system represented in the general state-variable form, you need also worry about controllability. if the system is not completely controllable, there are states (or some "mode" or linear combination of states) that are, effectively, not hooked up to the input $\mathbf{u}(t)$. now, that said, if the system is completely controllable, then every state in $\mathbf{x}(t)$ can be represented as the sum of components of that state that are driven by each input in $\mathbf{u}(t)$. so then first break it down to a single-input single-output question. and i'll try to answer that. $\endgroup$ Commented Dec 28, 2014 at 20:02
  • $\begingroup$ or is this already a single-input, single-output question? then it's a specific first-order system. perhaps i have acted prematurely in editing the question. $\endgroup$ Commented Dec 28, 2014 at 20:09
  • $\begingroup$ Either, matrix or scalar, same problem. A have to disagree with Sarwate's comment however. It can't be by definition otherwise I could state by definition, that y = x^2 is linear. A proof of linearity is required so that all the theorems than can be derived for LTI systems are valid. $\endgroup$
    – rhody
    Commented Dec 29, 2014 at 17:57

2 Answers 2


I'm going to assume the system has the initial condition $x(0) = 0$, as otherwise the system isn't linear.

Suppose the input $u_1(t)$ yields the output $x_1(t)$ and the input $u_2(t)$ yields the output $x_2(t)$.

We need to prove that the input $u(t) = \alpha u_1(t)+\beta u_2(t)$ yields the output $x(t) = \alpha x_1(t) + \beta x_2(t)$ for any scalars $\alpha$ and $\beta$.

The given input and output pairs must satisfy the differential equation and the initial condition. Hence, $x_1'(t) = Ax_1(t)+Bu_1(t)$ and $x_2'(t) = Ax_2(t)+Bu_2(t)$ for all times $t$ and $x_1(0) = x_2(0) = 0$.

Now, add $\alpha$ times the first equation to $\beta$ times the second equation to get: $\alpha x_1'(t) + \beta x_2'(t) = \alpha(Ax_1(t)+Bu_1(t)) + \beta(Ax_2(t)+Bu_2(t))$

Then, rearrange stuff to get:

$\dfrac{d}{dt}[\alpha x_1(t) + \beta x_2(t)] = A(\alpha x_1(t)+\beta x_2(t)) + B(\alpha u_1(t)+\beta u_2(t))$

$x'(t) = Ax(t)+Bu(t)$

Also, we have $x(0) = \alpha x_1(0)+\beta x_2(0) = \alpha \cdot 0 + \beta \cdot 0 = 0$.

Thus, the signals $u(t) = \alpha u_1(t)+\beta u_2(t)$ and $x(t) = \alpha x_1(t) + \beta x_2(t)$ satisfy the differential equation and the initial condition. Hence, the input $u(t) = \alpha u_1(t)+\beta u_2(t)$ yields the output $x(t) = \alpha x_1(t) + \beta x_2(t)$ for any inputs $u_1(t)$ and $u_2(t)$ and any scalars $\alpha$ and $\beta$.

Therefore, the system is linear, as desired.

  • $\begingroup$ Many thanks. I assumed that the initial conditions could be x(0) <> 0 hence my trouble in showing linearity. I assume then that many of the theorems in LTI theory that rely on linearity must make the same assumption that x(0) = 0, for example convolution? $\endgroup$
    – rhody
    Commented Dec 28, 2014 at 1:57
  • $\begingroup$ "I'm going to assume the system has the initial condition $x(0)=0$, as otherwise the system isn't linear. " i have to disagree with this, because $x(0)$ can simply be defined as $$ x(0) = \int\limits_{-\infty}^{0} u(\tau) h(-\tau) d\tau $$ and, even if $x(0) \ne 0$, if superposition applies, it's still linear. $\endgroup$ Commented Dec 28, 2014 at 20:16
  • $\begingroup$ Oh right, the system doesn't necessarily have to start operating at $t = 0$ or any particular finite starting time. If that is the case, just ignore everything regarding initial conditions. $\endgroup$
    – JimmyK4542
    Commented Dec 28, 2014 at 21:04

here is why, for a simple single-input, single-output (SISO) system, with no hidden states (i.e. completely controllable and completely observable), if superposition is known to be a valid property (called "additivity" by the math guys, you can also show that the scaling property (called "homogeneous of degree 1" by the math guys) of linearity is also the case for real and rational scalers.

you need to do a little hand-waving regarding continuity to extend this to the irrational scalers, because you can always get arbitrarily close to any irrational number with a rational one.

so let's say that $T\{ \cdot \}$ is something that transforms an input $x(t)$ into an output $y(t)$

$$ y(t) = T\{ x(t) \} $$

and this is true for all possible $x_1(t)$ and $x_2(t)$:

$$ T\{ x_1(t) + x_2(t) \} = T\{ x_1(t) \} + T\{ x_2(t) \} $$

now suppose $x_2(t) = x_1(t) = x(t)$

$$ T\{ x(t) + x(t) \} = T\{ x(t) \} + T\{ x(t) \} $$


$$ T\{ 2 \ x(t) \} = 2 \ T\{ x(t) \} $$

so you've just proved the linear scaling property for the real scaler $2$. how would you go about proving it for $3$? and then $4$? and then for any positive integer $N$? (it's called "proof by induction" and it's easy.)

so we're able to easily prove that superposition implies linear scaling for positive integers (and easily extended to all real integers):

$$ T\{ N \ x(t) \} = N \ T\{ x(t) \} \quad \quad N \in \mathbb{Z} $$

now how about this:

$$ T\left\{ \frac{1}{M} \ x(t) \right\} = ?? \quad \quad M \in \mathbb{Z} $$


$$ x_M(t) \triangleq \frac{1}{M} \ x(t) $$

so we know that

$$ T\left\{ M \ x_M(t) \right\} = M \ T\left\{ x_M(t) \right\} \quad \quad M \in \mathbb{Z} $$

but we also know that

$$ T\left\{ x_M(t) \right\} = T\left\{\frac{1}{M} \ x(t) \right\} $$


$$ T\left\{ M \ x_M(t) \right\} = T\left\{M \ \frac{1}{M} \ x(t) \right\} = T\{ x(t) \} $$

then we know that

$$ T\left\{ M \ x_M(t) \right\} = T\{ x(t) \} = M \ T\left\{ \frac{1}{M} \ x(t) \right\} \quad \quad M \in \mathbb{Z} $$


$$ T\left\{ \frac{1}{M} \ x(t) \right\} = \frac{1}{M} \ T\{ x(t) \} \quad \quad M \in \mathbb{Z} $$

so now we proved linear scaling for any reciprocal of an integer. sohow do you think we might do this form the general rational scaler $\frac{N}{M}$ where $N,M \in \mathbb{Z}$? i don't think it takes Einstein to see that

$$ T\left\{ \frac{N}{M} \ x(t) \right\} = \frac{N}{M} \ T\{ x(t) \} \quad \quad N,M \in \mathbb{Z} $$

so, just given superposition

$$ T\{ x_1(t) + x_2(t) \} = T\{ x_1(t) \} + T\{ x_2(t) \} $$

we know that scaling is also true

$$ T\{ \alpha x \} = \alpha \ T\{ x(t) \} \quad \quad \alpha = \frac{N}{M} \quad N,M \in \mathbb{Z}$$

and we know that

$$ T\left\{ \sum\limits_{i} \alpha_i x_i(t) \right\} = \sum\limits_{i} \alpha_i T \{ x_i(t) \} $$

where all $\alpha_i$ are rational. for the irrational $\alpha$, one must make an additional (and reasonable) assumption about the transform mapping $T\{\cdot\}$, that it is continuous, so that as rational $\alpha$ gets closer and closer to a irrational target (and changes very little), the behavior of $T\{\cdot\}$ will also change very little. i'm not gonna deal with that now.

but, from just a little thinking, it's clear that, at least for real and rational scalers, superposition implies linear scaling and then the whole of linearity in general. (says nothing about time-invariance, which is a whole 'nother property.)


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