Possible subspaces

Algebra Level 3

Let $V$ be the vector space $\mathbb{C}^2$ , or the set of vectors with two complex number components. Which of the following, if any, are subspaces of $V$ ?

The set of vectors $\begin{pmatrix}x\\y\end{pmatrix}$ for which $xy = 0$ .
The set of vectors $\begin{pmatrix}x\\y\end{pmatrix}$ for which $x$ and $y$ are both integers.

2, but not 1 Both 1 and 2 Neither 1 nor 2 1, but not 2

1 solution

Guillermo Templado
Jul 8, 2016

Relevant wiki: subspace

1.-

$A = \{(x,y) \in V = \mathbb{C}^2 \text{ / } xy = 0\}$ $(0,2) \in A\quad \wedge \quad (2,0) \in A$ but $(2,2) = (0,2) + (2,0)$ doesn't belong to $A \Rightarrow A$ is not a subspace of $V$ .

2.-

$B = \{(x,y) \in V \text{ / } \space \text{ x and y are both integers}\}$ $(1,1) \in B$ but $\frac{1}{2}(1,1) = (1/2, 1/2)$ doesn't belong to $B \Rightarrow B$ is not a subspace of $V$

why (2,2) is not a subspace? Thanks.

Pedro Valdericeda - 2 years, 4 months ago

because x=2 and y=2, xy=2*2 that is different from zero.

Alejandro Guevara - 1 year, 6 months ago

Shouldn't this problem specify the field which the vector space is over?

Sevatar Huang - 2 years, 4 months ago

@Pedro Valdericeda because 2*2 is not equal to zero and it was formed by linear combination of 2,0 and 0,2 which individually satisfied the properties of subspace but the linear combiantion is not closed under addition, hence not a subspace

Mathav Raj - 6 months, 3 weeks ago

Possible subspaces

1 solution

0 pending reports