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$$V=\left \{\begin{pmatrix}a & b\\ c & d\end{pmatrix} \mid a,b,c,d\in \mathbb{C} \text{ and } a+d\in \mathbb{R}\right \}$$

Why is this not a $\mathbb{C}$-vector space? Which of the following properties is not satisfied?

  1. $(V,+)$ is an abelian group, with the neutral element $0$.
  2. $\forall a, b \in K, \forall x \in V : (a + b) \cdot x = a \cdot x + b \cdot x$
  3. $\forall a \in K, \forall x, y \in V : a \cdot (x + y) = a \cdot x + a \cdot y$
  4. $\forall a, b \in K, \forall x \in V : (ab) \cdot x = a \cdot (b \cdot x)$
  5. $\forall x \in V : 1 \cdot x = x$ ( $1 = 1_K$ is the identity in $K$).

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EDIT:

If we want to check if it is a $\mathbb{R}$-vector space instead, the closures are satisfied, right?

The properties 2-5 are also satisfied, or not?

How can we check the property 1?

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How can we find a basis of $V$ as a $\mathbb{R}$-vector space?

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    $\begingroup$ Closure under scalar multiplication fails. For instance, certainly $I_{2} \in V$, where $I_{2}$ is the $2 \times 2$ identity matrix. But $i \cdot I_{2} \notin V$. $\endgroup$ Dec 3, 2016 at 3:01
  • $\begingroup$ So other than the 5 properties, the closure under the addition and the scalar multiplicatiom must hold? @AlexWertheim $\endgroup$
    – Mary Star
    Dec 3, 2016 at 3:03
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    $\begingroup$ Yes. Closure under addition follows from condition 1, for the record. $\endgroup$ Dec 3, 2016 at 3:04
  • $\begingroup$ Having at the defintion the following: "A $K$-vector space is set $V$ with an addition $V \times V \rightarrow V : (x, y) \mapsto x + y$ and a scalar multiplication $K \times V \rightarrow V : (\lambda , x) \mapsto \lambda \cdot x$, .... " this means that the closure under addition and scalar multiplication hold, right? @AlexWertheim $\endgroup$
    – Mary Star
    Dec 3, 2016 at 3:07
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    $\begingroup$ Yes. (characters) $\endgroup$ Dec 3, 2016 at 3:14

1 Answer 1

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By above conditions $V$ may be not closed to scalar multiplication. Let $\lambda=i$ then in $iA$ we will have $i(a+d)$ which can not belong to $\mathbb{R}$ because you assumption was $(a+d)\in \mathbb{R}.$ As a real vector space it will be $\mathbb{R}^4$ because $(a+d)\in \mathbb{R}$ is not an effective condition, that is always is true. It can be change to a condition like this $a+d=K$ a constant. in this case it will be a $3$-dimentional vector space.

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  • $\begingroup$ Ah ok... Thank you!! :-) $\endgroup$
    – Mary Star
    Dec 17, 2016 at 11:59

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