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I know that each operator in a $n$-dimensional complex space $A:E\to E$ is traingularizble. That is there exists a basis of the eigenvectors of $A$ wit which $A$ is triangular.

Now this claim would be true?

There are subspaces $F_i$ with $dim F_i=i$ which are invariant under $A$.

I think one can get the spaces $F_i$ generated by the eigenvectors $\{v_1,...,v_i\}$.

Any comments?

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The basis with respect to which the matrix is triangular does not necessarily consist of eigenvectors (in that case the matrix would be diagonalizable).

But the idea is correct. Let $\{v_1,v_2,\dots,v_n\}$ be a basis such that the matrix associated to $A$ is upper triangular, say $[b_{ij}]$, then \begin{align} Av_1&=b_{11}v_1 \in\operatorname{span}\{v_1\} \\ Av_2&=b_{12}v_1+b_{22}v_2 \in\operatorname{span}\{v_1,v_2\} \\ \vdots \\ Av_n&=b_{1n}v_1+b_{2n}v_2+\dots+b_{nn}v_n \in\operatorname{span}\{v_1,v_2,\dots,v_n\} \end{align} and so the subspaces listed are $A$-invariant with the prescribed dimensions.

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Yes, your observation is correct. Consider the space $V_j=span\{v_1,...,v_j\}$ where $v_i$ are eigenvectors $\forall i$. Then this subspace is an invariant subspace under $A$ as $Av_j=\sum_{i=1}^{j}A_{ij}v_i$ $\forall j$ as A is upper triangular. Hence you get a chain of subspaces ${0}=V_0\subset V_1\subset V_2\subset...\subset V_n=V$.

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