Is the number of entries $n$ of a vector $\vec v = \big(v_1, v_2, \ldots, v_n\big)$ in a vector space $V$ equal to $\dim(V)$? Is the number of entries $n$ of a vector $\vec v = \big(v_1, v_2, \ldots, v_n\big)$ in a vector space $V$ equal to $\dim(V)$? For example, take the vector space $P_3$, the set of polynomials of degree $3$ or less. It can be characterised by (or is isomorphic to?) the vector
$$\vec p = \big(a, b, c, d\big) \text{ for any polynomial } ax^3 + bx^2 +cx + d$$
Since the number of entries of $\vec p$ is $4$, does it follow that $\dim(P_3) = 4$. This question is more referring to the general case, not $P_3$, which I already know has dimension $4$.
 A: No. For example, if $V= \operatorname{sp} \{ (1,1) \}$, then $V$ has dimension one, but the number of elements in any point is two.
A: Not quite, as @paulinho pointed out. The dimension of a vector space V is determined by how many basis vectors span the space, that is, the minimum number of vectors from which any linear combination of them can create any vector in the field.
For $R^{n}$ and $P_{n-1}$ it is true that the number of entries in the vector corresponds to the dimension of the vector space but this is not true in general.
Consider the vector space corresponding to the x-axis on a coordinate plane. We can write all such vectors in the form of $\begin{pmatrix}{n}\\{0}\end{pmatrix}$ , but since they are spanned by the single vector $\begin{pmatrix}{1}\\{0}\end{pmatrix}$, the dimension of this space is $1$, which makes sense because the x-axis is a number line.

However, length isn't normally a term used for # of entries of a vector.
A: Generally, the dimension of a vector space is equal to the number of basis vectors (vectors that span the space).  
Here is an example of a vector space where the above does not hold:
$$\mathbf{0}_3 = (0,0,0)^T.$$  The vector space $\{(0,0,0)^T\}$ has dimension $0$.
