I have some questions to gain insigth on some core concepts of linear algebra:
Supose $V(\mathbb{C}^n)$ is a complex vector space, $\{v,u\}\in V$ are arbitrary distinct vectors with $v≠0$. Can we always find a linear operator $\tau:V\rightarrow V$ such that $\tau(v)=u$?
If it is the case, can we always find such an operator on a real vector space? Does this implies that every linear transformation has an eigenvalue?
Yes, it is always possible to find a map such that $v \mapsto u$, except when $v = 0$ because a linear transformation always maps the zero vector to itself. Indeed, you can take $v$ as the first basis vector of $V$ and maps it to $u$, such that the vector $u$ will appear as the first column of the matrix representation of $\tau$.
However, the eigenvalues of such a map are not necessarily real; indeed, the case $v = (1,0), u = (0,1) \in \mathbb{R}^2$ are linked by the linear operator $\tau = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, whose eigenvalues are $\pm i$.
If you want another concrete example, you can take $\tau = \frac{uv^\dagger}{\|v\|^2}$ for instance. The same map works for the real case when replacing $v^\dagger$ with $v^T$. Rotations constitute good candidates too.