Let $T:V\to V$ be a linear map on a vector space $V$ of dimension $n$. Assume that there exists $k\in \mathbb{N}$, such that for each $i$, $0\leq i\leq k$, there exists a subspace $V_i$ of $V$ (denoted $V_i \leq V$) with the following properties:
- $V_i \subset V_{i+1}$ for $0\leq i\leq k-1$
- $V_i \neq V_{i+1}$ for $0\leq i\leq k-1$
- $V_0 = \{0\}$ and $V_k = V$
- $TV_i \subseteq V_{i-1}$ for $1\leq i\leq k$
Then there exists a basis for $V$ such that the matrix of $T$ with respect to this basis is strictly upper triangular (i.e. all non-zero entries lie above the diagonal only)
So, I'm in the search of a basis $\mathcal{B} = \{v_1,…,v_n\}$ which makes $[T]_{\mathcal{B}}$ strictly upper triangular. This would mean that $Tv_j = \text{span}(v_1,v_2,…,v_{j-1})$. Intuitively, it makes sense to think about considering the basis of $V_i$ and extending it to a basis for $V_{i+1}$ (until we reach $V_n$), but I'm not sure how to put this down formally. Moreover, it's a little bothering that we don't know what $k$ is. Is that a typographical error in the question? Possibly not.
Please help me with hints so that I can work further in the right direction!
Best Answer
Hint
Take $\{e_1, e_2, \dots , e_{i_1}\}$ as a basis of $V_1$.
And by induction if $\{e_1, e_2, \dots, e_{i_1}, e_{i_1+1}, e_{i_1+2}, \dots, e_{i_{l-1}}\}$ is a basis of $V_{l-1}$, complete it into a basis $\{e_1, e_2, \dots, e_{i_1}, e_{i_1+1}, e_{i_1+2}, \dots, e_{i_{l-1}}, e_{i_{l-1}+1},e_{i_{l-1}+2},\dots, e_{i_l}\}$ of $V_l$.
At the last step and as $V_k = V$, you'll ensure that you get a basis of $V$.
Then look at $T(e_i)$ for $1 \le i \le n$ to prove that the matrix of $T$ in the built basis is strictly upper diagonal.