# Show that the identification $(a+ib)v= av+bTv$ defines a complex vector space

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?

• 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 $(1+i0)v$. Commented Jul 31, 2017 at 0:25
• @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.
– user173262
Commented Jul 31, 2017 at 0:28
• 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 $\mathbb C$. Commented Jul 31, 2017 at 0:33

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 $\;\mathbf R[X]/\ker \varphi\hookrightarrow\mathcal L(V)$.
• 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. Commented Jul 31, 2017 at 0:35
• 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. Commented Jul 31, 2017 at 0:44