Any linear transformation there exists a basis such that $\phi(v_i) = \sum_{j=1}^n a_{ij} v_j$.

Let $$V$$ be a finite dimensional vector space with basis $$\{v_1, \cdots, v_n\}$$ over an algebraically closed field $$K$$.

I want to prove the following theorem.

For any linear transformation $$\phi: V \rightarrow V$$, there exists a basis $$\{v_1, \cdots, v_n\}$$ of $$V$$ such that $$\phi(v_i) = \sum_{j=1}^n a_{ij} v_j$$, for $$a_{ij} \in K$$ with $$a_{ij}=0$$ whenever $$i>j$$.

Here is my trials:

Let $$v_j \in V$$, then since by construction $$\phi(v_j) \in V$$ can be written as $$\phi(v_j)$$ can be written as a basic element of $$V$$. Let $$\beta = \{v_1, \cdots, v_n\}$$. Then $$\phi(v_j) = [\phi(v_j)]_{\beta} [v]_{\beta} = \sum_{i} a_{ji} v_i$$. If we introduce inner product such that $$\langle v_i, v_j \rangle = \delta_{ij}$$, then I have $$\langle \phi(v_i), v_k \rangle =\langle \sum_ja_{ij} v_j, v_k \rangle = a_{ik}$$. But this does not give $$a_{ij} =0$$ whenver $$i>j$$...

This can be proved by induction. If $$n=1$$, the statement is trivial. Let $$n\in\Bbb N$$ and assume that the statement holds for any $$n$$-dimensional vector space. Let $$\phi$$ be a linear map from a vector space $$V$$ with dimension $$n+1$$ into itself. Since $$K$$ is algebraically closed, the characteristic polynomial of $$\phi$$ has some root in $$K$$; in other words, $$\phi$$ has some eigenvector $$v\in V$$. Let $$U$$ be a subspace of $$V$$ such that $$V=Kv\bigoplus U$$ and let $$\pi\colon V\longrightarrow U$$ be the projection from $$V$$ onto $$U$$ parallel to $$Kv$$ (that is, if $$\lambda\in K$$ and $$u\in U$$, then $$\pi(\lambda v+u)=u$$). Consider the map $$\psi\colon U\longrightarrow U$$ defined by $$\psi(w)=\pi\bigl(\phi(w)\bigr)$$. Then, by the induction hypothesis, there is some basis $$B=\{u_1,\ldots,u_n\}$$ of $$U$$ such that the matrix of $$\psi$$ with respect to $$B$$ is upper triangular. So, for each $$k\in\{1,2,\ldots,n\}$$, $$\phi(v_k)$$ is a linear combination of $$v,v_1,v_2,\ldots,v_k$$. In other words, if $$B_v=\{v,v_1,v_2,\ldots,v_k\}$$ then the matrix of $$\phi$$ with respect to $$B_v$$ is upper triangular.
Ok, so if this is too overkill/simple of an answer feel free to ignore me, but for any finite-dimensional vector space, $$V$$, over an algebraically closed field (this assumption is very important), your desired basis $$\{v_{1},\dots, v_{n}\}$$, that forces $$a_{ij} = 0$$ whenever $$i > j$$, always exists due to the fact that the matrix of any linear map that acts on such $$V$$ is similar to the Jordan Canonical Form. The basis for this is composed of the basis of each of the invariant subspaces corresponding to each Jordan Block of the Canonical Form.