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A) Is it true that always the group $\{(Z_1,Z_2) \in \mathbb{C}^2\;|\;Z_2 \in \mathbb{R}\}$ is a subspace of $\mathbb{C}^2$ as a vector space over $\mathbb{C}$?

B) Is it true that always the group $\{(Z_1,Z_2) \in \mathbb{C}^2\;|\;Z_2 \in \mathbb{R}\}$ is a subspace of $\mathbb{C}^2$ as a vector space over $\mathbb{C}$?

Thanks in advance.

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Is there any difference between (A) and (B)? –  Dylan Moreland Dec 11 '11 at 18:55
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It is a subspace over the field of real numbers but it is not a subspace over the field of complex numbers. [I assume you meant to say "a vector space over $\mathbb{R}$" in one of the questions.] –  Bill Cook Dec 11 '11 at 19:58
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Hint. Is your set closed under scalar multiplication by $i$? –  Arturo Magidin Dec 11 '11 at 20:05

1 Answer 1

up vote 1 down vote accepted

Let $W = \{ (z_1,z_2) \in \mathbb{C} \;|\; z_2 \in \mathbb{R} \}$. To conclude a subset of a vector space is a subspace, we need to verify 3 things:

  • The set is non-empty: This is obviously true (as is usually the case for such problems). But if we wish to be really picky...note that $(0,0) \in W$ so that $W$ is non-empty.

  • The set is closed under (vector) addition: This again is fairly obvious...suppose that $v,w \in W$. Then there exists $v_1,w_1 \in \mathbb{C}$ and $v_2,w_2 \in \mathbb{R}$ such that $v=(v_1,v_2)$ and $w=(w_1,w_2)$ (the second components are real). $v+w=(v_1,v_2)+(w_1,w_2)=(v_1+w_1,v_2+w_2) \in \mathbb{C}^2$. Also, notice that $v_2+w_2 \in \mathbb{R}$ because both $v_2$ and $w_2$ are in $\mathbb{R}$. Therefore, we have closure under addition.

  • The set is closed under scalar multiplication: This is true if your scalars are reals and false if they are complex numbers...suppose $r \in \mathbb{R}$ and $v=(v_1,v_2) \in W$ (so that $v_1\in\mathbb{C}$ and $v_2\in\mathbb{R}$). Then $rv=(rv_1,rv_2)$. $rv_1 \in \mathbb{C}$ since real times complex is still complex and $rv_2 \in \mathbb{R}$ since real times real is real. Thus $rv \in W$. On the other hand, $(0,1) \in W$ (since $0 \in \mathbb{C}$ and $1 \in \mathbb{R}$) but $i(0,1)=(0,i) \not\in W$ (since $i \not\in \mathbb{R}$). Thus $W$ is closed under real scalar multiplication but not complex scalar multiplication.

Therefore, $W$ is a subspace of $\mathbb{C}^2$ if we have dealing with real vector spaces, but $W$ is not a subspace if we are dealing with complex vector spaces.

By the way, in general, if $W$ is a subspace of $V$ as $\mathbb{K}$-vector spaces and $\mathbb{F}$ is a subfield of $\mathbb{K}$, then $W$ is a subspace of $V$ as $\mathbb{F}$-vector spaces. This is because subspaces are subgroups (under addition) regardless of the field of scalars and if $W$ is closed under scalar multiplication by elements in $\mathbb{K}$, then it is closed under scalar multiplication by any subset of $\mathbb{K}$ (the subspace property is preserved under restriction of the field of scalars). However, the converse is not true (we may lose closure under scalar multiplication if we try to enlarge our field of scalars) as this problem shows.

Also, notice that $W$ is isomorphic to $\mathbb{R}^3$ (as real vector spaces). It is 3-dimensional over the reals. However, if $W$ were to be a complex vector space it would have to be isomorphic to ??$\mathbb{C}^{1.5}$?? and have dimension ??3/2??. In general only real vector spaces of even dimension have any hope of also being complex vector spaces.

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