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?

  • 2
    $\begingroup$ 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. $\endgroup$ – logarithm May 30 at 14:57
  • $\begingroup$ I think it's also interesting to note that the dimension over $\mathbb{Q}$ is uncountable $\endgroup$ – Qidi May 30 at 15:13


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 $\mathbb{C}^2$ ?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.