Is asking about transition matrix in one dimensional random walk $X_n=U_0+U_1+...+U_n$ with $P(U_k=1)=p$, $P(U_k=-1)=1-p$ make sense?
If $n$ is even like $88$ then for example $p_{10,11}=P(X_n=11|X_{n-1}=10)$ is $0$ or is undefined? $X_{88}$ can't equal $11$ but $X_{87}$ couln't equal to $10$ in the first place.
If it make sense then this matrix has to be dependent on $n$ right? (be not homogeneous?)
The Markov chain is still time homogeneous - the transition matrix is still bi-diagonal.
The Markov chain cannot reach certain state in the odd / even steps does not mean it is not homogeneous. Some Markov chain can be periodic too.
Note the transition matrix stores the values of the transition probabilities $\Pr\{X_{k+1} = j | X_k = i\}$
By definition,
$$ \Pr\{X_1 = 1|X_0 = 0\} = p, \Pr\{X_1 = -1|X_0 = 0\} = 1 - p, \Pr\{X_1 = 0|X_0 = 0\} = 0$$
But these transition probabilities are different from marginal probabilities - i.e. we cannot simply say $\Pr\{X_1 = 0\} = 0$ without mentioning the initial distribution of $X_0$.
Also, these transition probabilities does not conditional on the value of $X_0$. i.e.
$$ \Pr\{X_3 = 0|X_0 = 0\} = 0 \neq p = \Pr\{X_3 = 0|X_2 = -1\} $$
You maybe still thinking of the probabilities like $$ \Pr\{X_3 = 0|X_2 = -1, X_0 = 0\} $$
This is true that this is ill defined because the event $X_2 = -1, X_0 = 0$ has a probability of $0$. But this quantity is not the same as the transition probability - the transition matrix just store information about something like $\Pr\{X_{k+1} = 0|X_k = -1\} $ without the information of the initial distribution $X_0$.