# Set of all $2*2$ matrix on $\mathbb Q$ is not a vector space over $\mathbb R$

why $V= \{\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}|a,b,c,d \in \Bbb Q\}$ is not a vector space over $\Bbb R$ under usual matrix addition and scalar multiplication. I check all 8 vector space axioms. but I couldn't find why this set is not a vector space.give me a hint.

• It isn't a vector space over R, but it is a vector space over Q, since the entries of the matrix must be rational. Jun 1, 2016 at 19:56

The set of $2\times2$ matrices over $\mathbb{Q}$ is not a vector space over $\mathbb{R}$ under any operations.

Indeed, the set is countable; on the other hand, if $V$ is a nontrivial vector space over $\mathbb{R}$ and $v\in V$, $v\ne0$, the map $\alpha\mapsto \alpha v$ is injective, proving that $|V|\ge|\mathbb{R}|$, so $V$ is uncountable.

The exercise is ill posed, though: the “usual scalar multiplication” is not defined.

• can you explain this answer in more detail? how the set $V$ is countable? Jun 1, 2016 at 20:14
• @Gune The product of two countable sets is countable; by induction, the product of a finite number of countable sets is countable. Jun 1, 2016 at 20:16

Hint:

if $A \in V$ and $\alpha$ is an irrational number than $\alpha \cdot A$ is not an element of $V$ because the product of a rational number with an irrational number is irrational. But, by definition, the scalar product must be a function $\cdot : \mathbb{R} \times V \to V$.