$f : V \rightarrow V$ : linear mapping
$ \text{dim}V = n$
Proposition;
If Invariant subspace $W_1, W_2, \cdots ,W_n(=V) \quad (\text{dim}W_k=k)$
$\quad s.t. \quad W_1 \subset W_2 \subset \cdots \subset W_n(=V)$ exist,
then $f$ has a triangular representation matrix.
Is this proof correct?
Proof
Because $\text{dim}W_1 = 1$, I let $\{ v_1 \}$ bese of $W_1$.
Because $\text{dim}W_2 = 2$ and $W_1 \subset W_2$, I can let $\{ v_1, v_2 \}$ base of $W_2$.
I repeat the same thing from $W_3$ to $W_n$,
I can let $\{ v_1, v_2, \cdots , v_n \}$ base of $W_n(=V)$.
Then, considering $f(W_k )\subset W_k$,
I can express
$ f(v_1)=c_{11} v_1 \\ f(v_2)=c_{12} v_1 + c_{22} v_2 \\ \quad \vdots \\ f(v_n) = c_{1n} v_1 + c_{2n} v_2 + \cdots + c_{nn} v_n$.
Then, representation matrix is
\begin{equation} \begin{pmatrix} c_{11} & c_{12} & c_{13} & \cdots & c_{1n} \\ 0 & c_{22} & c_{23} & \cdots & c_{2n} \\ 0 & 0 & c_{33} & \cdots & c_{3n} \\ \vdots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & c_{nn} \\ \end{pmatrix} \end{equation}