# Understanding the state vector in Quantum Computing

I have just started to learn QC. It is said that

The quantum state of $$N$$ qubits can be expressed as a vector in a space of dimension $$2^N$$

If there is $$1$$ qubit then we have two possible state vectors $$|0\rangle$$ and $$|1\rangle$$ or $$(0,1)$$ and $$(1,0)$$ respectively. Getting to $$2$$ qubits we have $$4$$ possible state vectors $$(1,0,0,0), (0,1,0,0), (0,0,1,0),$$ and $$(0,0,0,1)$$. Note that in each case, all entries are zero except 1. The point I am trying to get to is that:

1. $$2^N$$ seems like a big space but given a vector in this space - all components will be zero except $$1$$. So there are only $$2^N$$ possible values the state vector can take. Is this not correct? If not, why?

2. Why don't we say the space is $$N$$-dimensional. A $$N$$-bit string has $$2^N$$ possible values.

$${1 \over \sqrt{2}} |0\rangle + {1 \over \sqrt{2}} |1\rangle.$$
$${1 \over \sqrt{2}} (0,1) + {1 \over \sqrt{2}} (1,0) = (1/\sqrt{2},1/\sqrt{2}).$$
The vector space is not $$N$$-dimensional. The dimension of a vector space has a formal definition, and if you apply it, you will discover that the dimension of the vector space is $$2^N$$.