Is every real inner product vectorial space a complex one? Based on this question, I would like to know if every real inner product vectorial space is a complex one? I think yes, because $\mathbb R\subset \mathbb C$, so if a vector can be multiplied by a real scalar it can be multiplied by a complex one because every real number is a complex one.
 A: A vector space is defined as a quadruple $(V,\mathbb{K},+,\cdot)$ where $V$ is a set, $\mathbb{K}$ is a field, $+$ is a binary operation on $V$ such that $(V,+)$ is an abelian group, and $\cdot$ is an operation $\cdot: \mathbb{K}\times V \to V$ that satisfies some axioms. The kay fact here is that the  multiplication by a scalar ($a\cdot\vec v$) must be an element of $V$, so, if $V=\mathbb{R}^n$ The product of a complex number $a$ with a  vector $\vec v$ cannot be defined simply using the usual multiplication of a complex number for the components of the vector, because such multiplication gives a complex number so that the result is not an element of $\mathbb{R}^n$.
So the answer to your question is No, if you don't have some definition of the ''multiplication'' of a real number by complex numbers that gives as  result a real number and that satisfies the axioms of a vector space, or if you change in  a some nice way the set $V$ .
A: You can always complexify your space, i.e. take
$$V_\mathbb{C} := \mathbb{C}\otimes_\mathbb{R} V$$
(where you look at $\mathbb{C}$ as a $2$-dimensional real vector space. The vector space so obtained is naturally a complex vector space. Now extend the (real valued) inner product $\langle\cdot,\cdot\rangle_V$ into a complex valued one by asking
$$\langle z\otimes x,w\otimes y\rangle_{V_\mathbb{C}} = zw\langle x,y\rangle_V$$
for $z,w\in\mathbb{C}$. This is the only possible inner product on $V_\mathbb{C}$ restricting to $\langle\cdot,\cdot\rangle_V$ on $1\otimes V$.

In the other direction, there is the process of restriction of scalars: if you have a field morphism $\phi:k\to K$, then you can see every $K$-vector space as a $k$-vector space by defining a new scalar multiplication by
$$x\cdot v := \phi(x)\cdot v$$
for $x\in k$. In your question, you are considering the inclusion $\mathbb{R}\to\mathbb{C}$, but this tells you that you can see every complex vector space as a real one, and not the converse.
A: If you have a real vector space $V$ with an inner product $\cdot$, then you extend it to a complex vector space $W$ following the scheme:


*

*take the cartesian product $V\times V$. 

*define the multiplication by a scalar as
$$(a+ib)(x,y)=(ax-by,ay+bx)$$

*define the new scalar product $\bar\cdot$ as $$(x,y)\bar \cdot(v,w) = x\cdot v - y\cdot w+i(x\cdot w+y\cdot v)$$
This scheme allows to answer the question "If my space $V$ is a real one, how do I define a vector $iv$?"
