Initial conditions for systems described in state space - LTI or not? Suppose we have some system given by

$$\begin{aligned}
\dot{x}(t) &amp;= Ax(t) +Bu(t) \\
y(t) &amp;= Cx(t)+Du(t)
\end{aligned}$$

where $x(t)$ are the state variables, $y(t)$ is the output and $u(t)$ is the input. All matrices are constant. The same question applies to the discrete case

$$\begin{aligned}
x[n+1] &amp;= Ax[n] +Bu[n] \\
y[n] &amp;= Cx[n]+Du[n]
\end{aligned}$$

It is known that a system with non-zero initial conditions cannot be LTI. However, if $x(0)\neq0$, I don't see why the system above would not be LTI as it is expressed. As far as I know, if a system is expressed that way, it has to be linear and, as matrices do not depend on $t$, it should be time-invariant too.

So we have a system that has to be LTI as it is expressed in state space with constant matrices, but it can't be LTI because it has $x(0)\neq0$.

I can't see the mistake in the reasoning that leads me to this absurd contradiction. Can someone point it out?
If you look at Chapter 5 of:

https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-011-introduction-to-communication-control-and-signal-processing-spring-2010/readings/MIT6_011S10_notes.pdf

which is entitled "Properties of LTI State Space Models" , equation 5.33 doesn't seem to have a problem with initial conditions, or any other book (I stand corrected, there is one book) that I'm aware of.  Unless Oppenheim was touched with insanity, I'm inclined to accept his characterization that initial conditions don't disqualify a LTI system as "not linear" by his use of the term "zero input linear".

At the beginning of the notes, (and in Oppenheim and Shaefer 3rd edition) a LTI system is given as:

$$
y[n]=\sum_{k=-\infty}^{\infty} x[k] h[ n -k]  
$$
 which doesn't require $h[n]$ to be causal or stable.  $x[n]$ doesn't have to satisfy $x[n]=0 \;\text{for} \; n &lt;0$ .

There is emphasis  in the text that one needs to consider the entire history of $x[n]$, not just for $n \ge 0$.

let $$x[n] = \hat{x}[n]+\tilde{x}[n] $$
  where 
$$ \hat{x}[n] = \begin{cases} x[n]\; \text{for}\; n&lt;0 \; \text{and}\\
0 \; \text{for}\; n\ge 0 \end{cases}$$ 

and 
$$ \tilde{x}[n] = \begin{cases}0\; \text{for} \; n&lt;0 \; \text{and}\; \\ x[n] \; \text{for}\; n\ge 0 \end{cases}$$
 by linearity.

$$
y[n]=\sum_{k=-\infty}^{-1} \hat{x}[k] h[ n -k]  + \sum_{k=0}^{\infty} \tilde{x}[k] h[ n -k] 
$$
if $y[n]$ is causal,
$$
y[n]=\underbrace{\sum_{k=-\infty}^{-1} \hat{x}[k] h[ n -k]}_{\text{zero input linear}}  + \underbrace{\sum_{k=0}^{n} \tilde{x}[k] h[ n -k]}_{\text{zero state linear}}, \quad n \ge 0
$$

The essential point is that initial conditions account for prior input.  Where $n=0$ is referenced for $x[n]$ is arbitrary, which is another manifestation of time invariance.  Initial conditions are not some arbitrary value vexing the system.  if $x[n] = 0 $ for $n &lt;0 $ the initial conditions are zero.  

Let's try something else. Let $z[n]=\tilde{x}[n+1]$ (advance by one sample) and with $\tilde{x}[n]$, the system was LTI without controversy.
But now, 
$$
y[n]=\underbrace{ z[-1] h[n]}_{\text{zero input linear}}  + \underbrace{\sum_{k=0}^{n} z[k] h[ n -k]}_{\text{zero state linear}}, \quad n \ge 0
$$
and now we have an initial condition. A forward shift of 1 sample would make an LTI system nonlinear? 

The logical fallacy at the root of question is to use the definition of zero state linearity and applying it to the zero input case.