# Number of configurations, non-deterministic $LBA$ and $A_{LBA}$

The membership problem $$A_{LBA}$$ for a deterministic $$LBA$$ is decidable because the number of configurations that a $$LBA$$ can assume is finite. Since this number is also finite for a non-deterministic $$LBA$$, how can this information be used to show that the membership problem $$A_{LBA}$$ is decidable for a non-deterministic $$LBA$$?

P.S. Suppose the $$LBA$$ has $$|Q|$$ states, the size of the input is $$|w|$$ and the size of the alphabet is $$|\Gamma|$$. Then the number of possible configurations for this $$LBA$$ is $$|Q|*(|w|+2)*|\Gamma|^{|w|}$$. So it is finite.
There is a connection to Savitch's theorem which states that nondeterministic space can be simulated by deterministic space at the cost of squaring the space: $$\mathsf{NSPACE}(f(n)) \subseteq \mathsf{DSPACE}(f(n)^2)$$. Relevant here is that LBA are linear space TM's. Thus according to the theorem, a nondeterministic LBA can be simulated by a quadratic space deterministic TM. It is not known whether nondeterministic LBA can de determinized (meaning we do not need the squaring).