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Let $V$ be real $n$-dimensional vector space, and $T:V\to V$ is a linear map satisfying the condition $T^2(v)=-v$ for all $v \in V$. Then,

  1. Show that $n$ is an even integer.
  2. Use $T$ to make $V$ into a complex vector space such that the multiplication by complex numbers extends the multiplication by reals.
  3. Show that, with respect to the complex vector space structure on $V$ obtained in 2, $T:V\to V $ is a complex linear function.

This problem is bugging me for a while. And I have a few questions about it. I did no. 1 using the concept of minimal polynomials. [Another nice proof can be found here.] But the real troubles are question no. 2 and 3. The whole statement of Q.2 looks very vague to me. (For instance, I have doubts that, if I declare a real vector space to be a complex one, how can it be same as the previous one?) The follow up question has an equally dubious statement.

I would be glad if somebody takes the time to clarify what these two statements actually mean and exactly what I have to prove. Thank you.

[Source: This question can be found here (Question 25b).]

Community
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Sayantan
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2 Answers2

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Define complex scalar multiplication $ℂ×V→V$ by $(a+ib)v = av + b T(v)$. Prove that this defines a vector space. (The crucial point is that associativity law concerning scalar multiplication.) You also need to show that this really extends real multiplication, i.e. $av$ = $(a+0i)v$ where the left hand side term is to be understood as real multiplication as before. As an abelian group, $V$ is still the same, you just extended its “scalar field”. Then you can show that $T((a+ib)v) = (a+ib)T(v)$, meaning that $T$ is not only $ℝ$-linear, but also $ℂ$-linear.

k.stm
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    +1, just some remarks to show how nice this method is. You never need point 1., you get a complex vector space in 2. and this is always real-even-dimensional. For 3., you get commutation of $T$ with multiplication by $a+ib$ almost for free, because that multiplication is defined by a polynomial $a+bT$ in $T$. And if you identify $\Bbb C$ with $\Bbb R[X]/(X^2+1)$, then you get 2. almost for free from viewing $V$ as $\Bbb R[X]$-module, with $X$ acting as $T$, and observing that $X^2+1$ acts as $T^2+I=0$. – Marc van Leeuwen Jan 26 '13 at 10:41
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The point is indeed to “replace” ${\mathbb R}$ with ${\mathbb C}$, but better than just replace it, you want to lose nothing into the bargain. In other words, when you have the “dot” operations

$$ f_1 : {\mathbb R} \times V \to V \ \text{when }\ V \ \text{ is viewed as a real vector space} $$

$$ f_2 : {\mathbb C} \times V \to V \ \text{when } V \text{ is viewed as a complex vector space} $$

$f_1$ is the initial data of the problem, you cannot change it.

For $f_2$, you could choose anything you like a priori, but it is more interesting if $f_2$ extends $f_1$ (i.e. $f_1(r,v)=f_2(r,v)$ whenever those two terms make sense). This is what K. Stm’s answer does.

Ewan Delanoy
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