# dimension of a vector space over $\mathbb{C}, \mathbb{R}$ and $\mathbb{Q}$

This problem is taken from Linear Algebra by Fueng Zhang. R is real numbers, C is complex numbers and Q is rational numbers.

Let $$V = \{(x,y) \mid x,y \in \mathbb{C}\}$$. Under the standard addition and scalar multiplication for ordered pairs of complex numbers, is $$V$$ a vector space over $$\mathbb{C}$$? Over $$\mathbb{R}$$? Over $$\mathbb{Q}$$ If so, find the dimension of $$V$$.

The answer key says:

Yes, over $$\mathbb{C}, \mathbb{R}$$ and $$\mathbb{Q}$$. The dimensions are $$2,4,\infty$$, respectively.

I do understand that the dimension over $$\mathbb{C}$$ is 2 and over $$\mathbb{R}$$ is 4, but how come the dimension over $$\mathbb{Q}$$ is infinite?

• There are numbers like $e$ (transcendental numbers)that are not roots of any polynomial with rational coefficients. Therefore $1,e,e^2,e^3,...$ are linearly independent over $\mathbb{Q}$. If they were dependent then there would be some rational numbers $r_1,...,r_n$, such that $r_1e^{m_1}+...+r_ne^{m_n}=0$, which is a polynomial with rational coefficients of which $e$ is a solution. – logarithm May 30 at 14:57
• I think it's also interesting to note that the dimension over $\mathbb{Q}$ is uncountable – Qidi May 30 at 15:13

Hint:

Prove that any finite dimensional vector space over $$\mathbb Q$$ (or any other countable field) is countable.

Suppose that $$\mathbb{C}^2$$ is a finite dimensional $$\mathbb{Q}$$ vector space, then

$$\mathbb{C}^2 \cong \mathbb{Q}^n$$ for some $$n \geq 1$$.

In particular, it would follow that $$\mathbb{C}^2$$ is countable. Contradiction.

The other two are easy: you can just write down bases for them

For example, $$\{1,i\}$$ is $$\mathbb{R}$$-linearly independent over $$\mathbb{C}$$. What is an $$\mathbb{R}$$-basis for $$\mathbb{C}^2$$ ?