Invariant subspace of all dimensions. Let $V$ be a finite dimensional vector space over $ \Bbb C,$ and suppose that $\dim(V) = n$. Prove that if
$T \in L(V)$ then for each $k$ with $0 \leq k \leq n$, $T$ has an invariant subspace of dimension $k.$
First of all it seems clear to me that we wish to use induction its just not clear to me on what. originally i tried using induction to say it was true when $n=0,1$ trivially assume its true for $1<k<n $ and tried to show it was true for $n$ by looking at the eigenspaces because we know these are invariant but it turns out this is not the right approach the problem is although the may work for some lower values we cant actually say it works for n because we may not have enough eigenvectors to span all of $n$ (it works when the map is diagonalizable only). i did try a non-inductive approach but it was more complicated also appeared to not work.
 A: In the comments, you thought about using upper triangular matrices. Try proving the following:

An operator $T\in L(V)$ can be represented with a triangular matrix with respect to a certain basis of $V$ iff there exists a basis $(v_1,\dots,v_n)$ such that $\forall 0\le k\le n,\,\text{span}(v_1,\dots,v_k)$ is an invariant subspace of $T$.

A: Hints for another approach without triangular form:
The characteristic polynomial of $\;T\;$ has all its roots in $\;\Bbb C\;$, say $\;a_1,...,a_k\;$ . Induction on $\;n\;$ : for $\;n=1\;$ the claim is trivial as the trivial space and the whole space are $\;T\,-$ invariant.
Assume for $\;\dim V<n\;$ and we prove for $\;\dim V=n\;$ . Take the eigenspace $\;V_1\;$ related to the eigenvalue $\;a_1\;$ . Of course, $\;\dim V_1\ge1\;$, so the quotient space $\;V/V_1\;$ has dimension less than $\;\dim V=n\;$ . We also have a well-defined linear operator (check all this!)
$$T'\in\mathcal L\left(V/V_1\right)\;,\;\;T'(v+V_1):=Tv+V_1$$
Well, now apply induction and do some mathematics...
