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This is the first part of the exercise 17 in page 286 of Linear algebra done right, third edition of Axler.

Suppose that $V$ is a real finite-dimensional vector space and $T\in\mathcal L(V)$ satisfies $T^2=-I$. Define a complex scalar multiplication on $V$ as follows: if $a,b\in\Bbb R$ then

$$(a+ib)v=av+bTv$$

(a) Show that the scalar multiplication defined above and the addition on $V$ makes $V$ a complex vector space

The question is: I dont know exactly what I must do here. What I did was show that the set of operators

$$H:=\{aI+bT:a,b\in\Bbb R\}\subseteq\mathcal L(V)$$

behave as the field $\Bbb C$, that is, each $j\in H\setminus\{0\}$ have a multiplicative inverse, a negative, a "conjugate" and in general it have the structure of a field like $\Bbb C$.

Also I had shown that $V_T$ (the vector space $V$ redefined under the above scalar multiplication) is a vector space, and that $av= bTv$ if and only if $a=b=0$ or $v=0$.

This is enough to say that $V_T$ is a complex vector space? There is other way to do this exercise?

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    $\begingroup$ You don't actually need to show those properties of $H$. You just need to show the vector space axioms. The ones involving only addition are immediate because $V$ is a real vector space, so essentially what's left is $z(wv)=(zw)v$, $1v=v$, $z(u+v)=zu+zv$, $(z+w)v=zv+wv$ for $u,v\in V$, $z,w\in\mathbb C$. Note that "$1v$" is interpreted as $(1+i0)v$. $\endgroup$
    – stewbasic
    Commented Jul 31, 2017 at 0:25
  • $\begingroup$ @stewbasic what confused me was the phrase "makes $V$ a complex vector space", instead of just "makes $V$ with the above scalar multiplication a vector space". So basically I must shows that $V_T$ is a vector space. $\endgroup$
    – user173262
    Commented Jul 31, 2017 at 0:28
  • $\begingroup$ Your second statement is more formally accurate. It is convenient and common (but less accurate) to say "$V$ is a vector space" rather than "$(V,+,\cdot)$ is a vector space" (where $+:V\times V\to V$ and $\cdot:K\times V\to V$). Here "complex vector space" just means vector space over $\mathbb C$. $\endgroup$
    – stewbasic
    Commented Jul 31, 2017 at 0:33

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*Hint:**

Consider the $\mathbf R$-algebra homomorphism: \begin{align} \varphi\colon\mathbf R[X]&\longrightarrow \mathcal L(V)\\ X&\longmapsto T \end{align} Determine $\ker \varphi$ and use the induced injection $\;\mathbf R[X]/\ker \varphi\hookrightarrow\mathcal L(V)$.

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  • $\begingroup$ This seems to answer the question "How do I show $H\cong\mathbb C$?". But the OP has shown this and is asking whether the exercise is actually asking for this. $\endgroup$
    – stewbasic
    Commented Jul 31, 2017 at 0:35
  • $\begingroup$ Well, that's the way I would do it, as there are very few verifications to do. It then becomes trivially a $\mathbf C$-vector space. $\endgroup$
    – Bernard
    Commented Jul 31, 2017 at 0:44

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