# If each complex operator has invariant subspaces of any dimension?

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 $$\{v_1,...,v_i\}$$.

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.
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 $${0}=V_0\subset V_1\subset V_2\subset...\subset V_n=V$$.