Complexification of a Vector Space
Introduction
Suppose we are given a finite-dimensional $\mathbb{R}$-vector space $V$ and a linear operator $A : V \to V$ on it. Finding the directions in which the action of $A$ becomes simplest, namely a scalar multiple, that is, finding a scalar $\lambda$ and $\mathbf{x} \ne \mathbf{0}$ satisfying the following, is the eigenvalue problem.
$$ A \mathbf{x} = \lambda \mathbf{x} $$
Since $V$ is finite-dimensional, $A$ corresponds to a matrix $\left[ A \right]$, and since the scalar field is the real numbers, this matrix has real entries. The eigenvalues of the matrix $[ A ]$ can be obtained by solving the characteristic polynomial, and because $[ A ]$ has only real entries, the characteristic polynomial is a polynomial with real coefficients. However, since the field of real numbers is not algebraically closed, there are cases in which the roots of the characteristic polynomial do not exist in the reals, in other words, the eigenvalues cannot be found within the real numbers. This means that diagonalization is impossible, and that tools such as eigenvalue decomposition and spectral analysis cannot be fully used over the field of real numbers. Since the field of complex numbers is an extension field of the field of real numbers, simply enlarging the scalar field to the complex numbers allows us to obtain all eigenvalues and eigenvectors. Carrying this out for the entire vector space is called complexification.
Definition
For a finite-dimensional $\mathbb{R}$-vector space $V$ and the 2-dimensional $\mathbb{R}$-vector space $\mathbb{C}$, the complexification $V_{\mathbb{C}}$ of $V$ is defined as follows.
$$ V_{\mathbb{C}} := V \otimes_{\mathbb{R}} \mathbb{C} $$
Here, $\otimes_{\mathbb{R}}$ is the tensor product of $V$ and $\mathbb{C}$ regarded as $\mathbb{R}$-vector spaces. Since $\mathbb{C}$ is $2$-dimensional when viewed as an $\mathbb{R}$-vector space, $V \otimes_{\mathbb{R}} \mathbb{C}$ is an $\mathbb{R}$-vector space of dimension $2 \dim V$. Equipping it with complex scalar multiplication as below makes it a $\mathbb{C}$-vector space.
$$ \alpha \cdot (\mathbf{v} \otimes z) := \mathbf{v} \otimes (\alpha z), \qquad \alpha, z \in \mathbb{C} $$
Simple Definition
For a finite-dimensional $\mathbb{R}$-vector space $V$, the following set is called the complexification $V_{\mathbb{C}}$ of $V$.
$$ V_{\mathbb{C}} = \left\{ \mathbf{v} + i\mathbf{w} : \mathbf{v}, \mathbf{w} \in V \right\} $$
Addition and scalar multiplication are defined respectively as follows.
$$ (\mathbf{v}_{1} + i\mathbf{w}_{1}) + (\mathbf{v}_{2} + i\mathbf{w}_{2}) := (\mathbf{v}_{1} + \mathbf{v}_{2}) + i(\mathbf{w}_{1} + \mathbf{w}_{2}) $$
$$ (a + bi)(\mathbf{v} + i\mathbf{w}) := (a\mathbf{v} - b\mathbf{w}) + i(a\mathbf{w} + b\mathbf{v}) $$
Here, the $i$ in $\mathbf{v} + i\mathbf{w}$ is merely a formal symbol for writing two vectors side by side, and the scalar multiplication is defined so that $i^{2} = -1$ holds for this symbol.
Explanation
The two definitions describe the same space. Since the $\mathbb{R}$-basis of $\mathbb{C}$ is $\left\{ 1, i \right\}$, every element of $V \otimes_{\mathbb{R}} \mathbb{C}$ can be written as a sum of elements of the form $\mathbf{v} \otimes 1$ and $\mathbf{v} \otimes i$, and it suffices to make the following correspondence.
$$ \mathbf{v} \otimes 1 \longleftrightarrow \mathbf{v} + i\mathbf{0}, \qquad \mathbf{v} \otimes i \longleftrightarrow \mathbf{0} + i\mathbf{v} $$
In general, the correspondence is as follows.
$$ \mathbf{v} \otimes (a + bi) \longleftrightarrow a\mathbf{v} + i(b\mathbf{v}) $$
Dimension
Let $\left\{ \mathbf{e}_{1}, \dots, \mathbf{e}_{n} \right\}$ be an $\mathbb{R}$-basis of $V$. If $\mathbf{v} = \sum_{k} a_{k}\mathbf{e}_{k}$ and $\mathbf{w} = \sum_{k} b_{k}\mathbf{e}_{k}$, then an element of $V_{\mathbb{C}}$ is uniquely expressed as follows.
$$ \mathbf{v} + i\mathbf{w} = \sum_{k=1}^{n} (a_{k} + i b_{k}) \mathbf{e}_{k} $$
That is, a real basis of $V$ becomes, as it is, a complex basis of $V_{\mathbb{C}}$, so the following holds.
$$ \dim_{\mathbb{C}} V_{\mathbb{C}} = \dim_{\mathbb{R}} V = n, \qquad \dim_{\mathbb{R}} V_{\mathbb{C}} = 2n $$
Complexification of Linear Operators
For an $\mathbb{R}$-linear transformation $A : V \to V$, the map $A_{\mathbb{C}} : V_{\mathbb{C}} \to V_{\mathbb{C}}$ defined as below is called the complexification of $A$.
$$ A_{\mathbb{C}}(\mathbf{v} + i\mathbf{w}) := A\mathbf{v} + i(A\mathbf{w}) $$
That $A_{\mathbb{C}}$ is $\mathbb{C}$-linear can be verified directly by following the definition. Moreover, if the real basis $\left\{ \mathbf{e}_{1}, \dots, \mathbf{e}_{n} \right\}$ of $V$ is used as it is as a complex basis of $V_{\mathbb{C}}$, then $A_{\mathbb{C}}\mathbf{e}_{k} = A\mathbf{e}_{k}$, so the matrix representation with respect to this basis is the same as that of $A$.
$$ \left[ A_{\mathbb{C}} \right] = \left[ A \right] $$
Hence the characteristic polynomials of the two operators are also the same. The only difference is the field over which the polynomial is solved, and now, by the fundamental theorem of algebra, roots always exist, so $A_{\mathbb{C}}$ always has eigenvalues.
Since the characteristic polynomial has real coefficients, non-real eigenvalues come in conjugate pairs. The same goes for eigenvectors: since $A_{\mathbb{C}}$ commutes with conjugation, taking the conjugate of both sides of $A_{\mathbb{C}}\mathbf{z} = \lambda\mathbf{z}$ gives the following.
$$ A_{\mathbb{C}}\overline{\mathbf{z}} = \overline{A_{\mathbb{C}}\mathbf{z}} = \overline{\lambda\mathbf{z}} = \overline{\lambda}\overline{\mathbf{z}} $$