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It's an exercise of the book Linear Algebra Done Right. I'm not clear about how to prove these problems, would you please offer me some suggestion about how to improve this kind of ability, thanks a lot.

When
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2 Answers2

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A vector space is a non-empty set $V$ equipped with "operations" of addition ($+$) and scalar multiplication ($\cdot$). A non-empty subset $W$ is a subspace if $W$ is a vector space under the operations of $V$. This turns out to be equivalent to requiring that $W$ is closed under addition and closed under scalar multiplication.

Here, $V$ is the set $\mathbf{C}^{2}$ of ordered pairs of complex numbers, $W$ is the set $\mathbf{R}^{2}$ of ordered pairs of real numbers, and the fact that you've referred to "the complex vector space" all but asserts that addition means componentwise addition, and scalar multiplication means componentwise multiplication by a complex number: $$ (z_{1}, z_{2}) + (w_{1}, w_{2}) = (z_{1} + w_{1}, z_{2} + w_{2}),\qquad c(z_{1}, z_{2}) = (cz_{1}, cz_{2}). $$

To determine whether $W$ is a subspace of $(V, +, \cdot)$, the questions to ask yourself are:

  1. Is $W$ closed under addition? That is, if $(z_{1}, z_{2})$ and $(w_{1}, w_{2})$ are pairs of real numbers, is their sum? (Answer: Yes. Why?)

  2. Is $W$ closed under scalar multiplication? That is, if $(z_{1}, z_{2})$ is a pair of real numbers, and if $c$ is an arbitrary complex number, is $(cz_{1}, cz_{2})$ a pair of real numbers? (No: Take any vector other than $(0, 0)$ and any non-real scalar $c$.)

Since $W$ is not closed under scalar multiplication, $W$ is not a subspace of the complex vector space $(\mathbf{C}^{2}, +, \cdot)$.

Any time you deal both with complex vector spaces and real vector spaces, you have to be certain of what "scalar multiplication" means. For example, the set $\mathbf{C}^{2}$ is also a real vector space under the same addition as before, but with multiplication only by real scalars, an operation we might denote $\cdot_{\mathbf{R}}$.

The real vector space $(\mathbf{C}^{2}, +, \cdot_{\mathbf{R}})$ is a different vector space than the complex vector space $(\mathbf{C}^{2}, +, \cdot)$, and $\mathbf{R}^{2}$ is a subspace of $(\mathbf{C}^{2}, +, \cdot_{\mathbf{R}})$, as you can easily check by mimicking the preceding proof.

Andrew D. Hwang
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  • So when checking for closure under scalar multiplication, the scalar factor is supposed to be a member of $\textit{V}$, that is: $\textit{c} \in \textit{V}$ and not $c\in \textit{W}\ \cap \textit{V}$? – Labbsserts Feb 16 '21 at 17:20
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The vector $\mathbf v=(1,1)$ is in $\Bbb R^2$ but $i\mathbf v$ is not.

ajotatxe
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  • I think the ability of finding a counter-example is what I'm lacking T_T – When May 04 '15 at 00:32
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    If the question is whether $\Bbb R^2$ is a subspace of a complex vector space, the scalars are the complex numbers. Otherwise the question would not make sense at all. – ajotatxe May 04 '15 at 00:36
  • @ajotatxe I suppose I misunderstood how the OP was thinking of $\mathbb{R}^2\subset \mathbb{C}^2$; I was making the identification $\mathbb{C}=\mathbb{R}^2$ and imagining $\mathbb{R}^2$ to be embedded in $\mathbb{R}^4=\mathbb{C}^2$ as the set of vectors of the form $(a,b,0,0)$ with $a,b\in\mathbb{R}$. However, I see now what was instead meant was $\mathbb{R}\subset \mathbb{C}$, so that with my original identification we'd actually have $\mathbb{R}^2={(a,0,b,0) \mid a,b\in\mathbb{R}}$ (granted, $\mathbb{R}$ here is being used in different ways, which is rather unfortunate...) – Hayden May 04 '15 at 00:43