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 $i>j$…
Best Answer
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.