In Halmo's book 'Finite dimensional vector spaces' there's a question I'm kind of stuck on in chapter 1.

$1 (b)$ Every complex vector space $\cal V$ is intimately associated with a real vector space $\cal V^-$; the space $\cal V^-$ is obtained from $\cal V$ by refusing to multiply vectors of $\cal V$ by anything other than real scalars. If the dimension of the complex vector space $\cal V$ is $n$, what is the dimension of the real vector space $\cal V^-$?

My thoughts:

$\cal V$ has dimension $n$ so we can find a basis of $n$ vectors $\{v_1,\ldots, v_n \}$. These vectors span the space any combination:

$\sum\limits_{k=1}^n \alpha_k v_k \neq 0$ unless all the $\alpha_k =0$ with $\alpha_k \in \Bbb C$. This also holds for the subset of real numbers in $\Bbb C$ so the same vectors must be linearly independent as a real vector space.

Thus the dimension of $\cal V^-$ must be at least $n$.

If $\cal V^-$ is a subspace of $\cal V$ then the spaces are equal, otherwise I would just assume it's dimension is $n$.

If it were more than $n$ I could just consider $n+1$ vectors in $\cal V $ and then they'd be linearly dependent.

This just feels very unintuitive if I've gone about it correctly.

  • $\begingroup$ "If it were more than n I could just consider n+1 vectors in V and then they'd be linearly dependent." $\mathbb{R}$-linearly or $\mathbb{C}$-linearly? $\endgroup$ – InTransit Jul 31 '14 at 7:52
  • $\begingroup$ @intransit Yeh good point. It would be $\Bbb C$ linearly dependent. I'm stuck with that. But if the dimension went up that would even less intuitive to me... $\endgroup$ – snulty Jul 31 '14 at 12:36

Hint. Note that linearly (in)depence depends on the scalar field. As an example, lets consider $V = \mathbb C$. Then $V$ is obviously an $1$-dimensional vector space over $\mathbb C$ and $v_1 = 1$ and $v_2 = i$ are linearly depended (for the scalars $\lambda_1 = i$ and $\lambda_2 = -1$ we have $\lambda_1 v_1 + \lambda_2 v_2 = 0$). But over $\mathbb R$ $v_1$ and $v_2$ are in(!)dependent: Let $\lambda_1, \lambda_2 \in \mathbb R$ with $\lambda_1 v_1 + \lambda_2 v_2 = 0$, then $\lambda_1 + \lambda_2 i = 0$, as $\lambda_i$ are real, that can only hold iff $\lambda_1 = \lambda_2 = 0$. So $V^-$ has $\mathbb R$-dimension (at least) two.
Hint for the general case: Consider $\{v_1, \ldots, v_n, iv_1, \ldots, iv_n\}$ for a $\mathbb C$-basis $\{v_1, \ldots, v_n\}$ of $V$. Show that is an $\mathbb R$-basis.

  • $\begingroup$ First Thanks for the response. But how do I know that the vectors $\{i v_k\mid k \in \{1,\ldots, n\}\}$ are in $\cal V^-$? Like in the case that the vector space was $\Bbb C$, that made sense for me but in general...? $\endgroup$ – snulty Jul 31 '14 at 12:57
  • 1
    $\begingroup$ Note that $V^-$ has the same underlying set as $V$. For a vector $v \in V$, multiplication by $i$ is well-defined (as $V$ is a $\mathbb C$-vector space), so $iv \in V$. But as sets $V$ and $V^-$ are the same. $\endgroup$ – martini Jul 31 '14 at 13:02
  • $\begingroup$ Right Yeh. I think I see what you mean. You think of the entire space $\can V$ but with future operations you only allow ones with real scalars. So dimension can change as can linear dependence of sets. Am I understanding correctly? $\endgroup$ – snulty Jul 31 '14 at 16:19
  • $\begingroup$ Also so the answer in general would be $2n $ I believe then is what is implied. $\endgroup$ – snulty Jul 31 '14 at 16:37
  • $\begingroup$ @snulty Yes, you are right. $\endgroup$ – martini Aug 1 '14 at 6:41

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