# the set of all complex numbers constitutes a one-dimensional complex vector space.

Show that the set of all real numbers, with the usual addition and multiplication, constitutes a one-dimensional real vector space, and the set of all complex numbers constitutes a one-dimensional complex vector space.

I know that all properties to be vector space are fulfilled in real and complex but I have difficulty is in the dimension and the base of each vector space respectively. Scalars in the vector space of real numbers are real numbers and likewise with complexes? The basis for both spaces is $\{1\}$ or for the real ones it is $\{1\}$ and for the complexes it is $\{i\}$? How does one prove that these spaces have dimension $1$? Thank you very much.

The set of real numbers is one dimensional over the reals (meaning our scalars are real) and the set of complex numbers is one dimensional over the complex numbers (meaning that our scalars are complex).

Why is $\{1\}$ a basis for each?

Well, consider any real number $x$. Then we simply choose the scalar $x$ and basis element $1$ to have $$x = x \cdot 1.$$

Similarly, for any complex number $z$, choose the complex scalar $z$ and basis element $1$ to get $$z = z \cdot 1.$$

Then it follows that $\{1\}$ spans $\mathbb{R}$ in the first case and it spans $\mathbb{C}$ in the second.

I leave it to you to see that a singleton set is always linearly independent (assuming the Singleton is non-zero).

Does this make sense?

Note: There are infinitely many bases for each vector space but the important thing is that every basis will have the same cardinality. In this case, any basis will consist of $1$ element. If we take the complex numbers with scalars from the complex plane then we could take $\{i\}$ as a basis since for any $z \in \mathbb{C}$, we have $$z = (-i z) \cdot i .$$ So, $\{i \}$ spans $\mathbb{C}$ in this case.

When the question asks you to show that it is a 'one-dimensional complex vector space', it implies that the set of scalars for scalar multiplication is $\mathbb{C}$.

Now, it doesn't make much sense to speak of the basis, but rather of a basis. $\{1\}$ is a basis for the real vector space $\mathbb{R}$, as is $\{x\}$ whenever $x\in\mathbb{R}\setminus\{ 0\}$. With this in mind, is $\{1\}$ a basis for the complex vector space $\mathbb C$? Is $\{i\}$?

Finally, the dimension of a vector space $V$ is the cardinality of elements (ie, how many elements) in a basis of $V$. This is well defined, because one can show that any two bases for $V$ have the same cardinality.

In order to show that $\mathbb{C}$ has complex dimension $1$, you need to exhibit a basis -- that is, a linearly independent set that spans $\mathbb C$ -- which has a single element.