The difference equation is given \begin{align*} x(k + 1) = Ax(k) + Bu(k) \end{align*} with an initial condition $x(0)= x_0$. Inductively, I derived the solution
$$x(k) = A^kx_0+\sum_{j=0}^{k-1} A^{k-j-1}Bu_{j},$$ which can be also find in this question - Derivation of a formula for discrete time case and probably in much difference-equation stuff.
Now, I want to show that the solution is unique.
My idea: I can prove that $A$ is nonsingular matrix and so $\Phi(k,0) = A^{k}$ is a fundamental matrix of the system above. And we define $\Phi(k,j) = A^{k-j}$.
Any solution of homogenous system $$x(k + 1) = Ax(k) \\ x(0)= x_0$$ can be written as $x(k) = \Phi(k,0)c$, where $\Phi$ is a fundamental matrix and $c$ is some constant vector. Then I can show that $y(k) = x(k) - x_p(k) = A(k-1)y(k-1)$, where $x_p(k)$ is a particular solution of the non-homogenous system. Hence $y(k)$ is a solution of homogenous system and it can be expressed as $$y(k) = x(k) - x_p(k) = \Phi(k,0)c.$$
Then I show that particular solution can be expressed as
$$x_p(k) = \sum_{j=0}^{k-1} \Phi(k,j+1)Bu_{j}$$ with $x_p(0) = 0$. Putting these two equation together one has
\begin{align*} x(k) &= \Phi(k,0)c + x_p(k) \\ &=\Phi(k,0)c + \sum_{j=0}^{k-1} \Phi(k,j+1)Bu_{j}. \end{align*}
Which is still not unique, but we have the initial condition
\begin{align*} x(0) &=\Phi(0,0)c \implies c = x_0. \end{align*}
And this completes the proof.
Question: Is this really necessary when i can derive whole equation inductively? Is nonsingularity of $A$ a necessary condition of uniqness?
Thank you in advance.
Unlike in the case of continuous time, for difference equations the solution is unique (in your particular case) simply because you were able to find an explicit formula that gives the solution.
In other words, you are completely right when you ask "Is this really necessary when i can derive whole equation inductively?" The answer in "no". Moreover, no condition on $A$ is necessary.