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

Question

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 $i>j$.

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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 $i>j$...

Answer 1

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.

Answer 2

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.

P.S. I will be following this post to see if there's a different, more clever proof of this fact!