Show that the operator has a chain of invariant subspaces. 
Let $V$ be a $n$-dimensional vector space over $\Bbb C$ and let $T:V\rightarrow V$ be any linear operator. Show that  $T$ has a chain $V_0\subseteq V_1\subseteq\ldots \subseteq V_n=V$ of invariant subspaces such that $\dim V_i=i$ for $0\le i\le n$.

Here a subspace $U$ of $V$ is an invariant subspace of $V$ if $T(U)⊆U$ and in this case $T|_U:U→U$ is a linear operator on $U$.
Please help me to solve this.
 A: For an elementary argument not requiring the Jordan canonical form: we argue by induction on $n$.  For the base case $n=0$, the statement is trivial: the required chain is $V_0 = \{ 0 \} = V$.  (If you prefer to let the base case be $n=1$, then the chain there is $\{ 0 \} \subsetneq V$.)
Now, suppose $n \ge 1$.  Then since $\mathbb{C}$ is algebraically complete, $T$ has at least one eigenvector; so let $x \ne 0$ be an eigenvector.  Since $T x = \lambda x$ for $\lambda$ the corresponding eigenvalue, we see that $\langle x \rangle$ is an invariant subspace of $T$.  Now, consider the induced operator on the quotient space, $\bar T : V / \langle x \rangle \to V / \langle x \rangle$.  This is a linear operator on an $n-1$-dimensional subspace, so by inductive hypothesis, we can find a chain $V_0 \subsetneq V_1 \subsetneq \cdots \subsetneq V_{n-1} = V / \langle x \rangle$ of $\bar T$-invariant subspaces.
Now, if $\pi : V \to V / \langle x \rangle$ is the projection operator, then we conclude that
$$ \{ 0 \} \subsetneq \pi^{-1}(V_0) \subsetneq \pi^{-1}(V_1) \subsetneq \cdots \subsetneq \pi^{-1}(V_{n-1}) = V $$
is a chain satisfying the requirements.
A: I assume that you know the Jordan canonical form. So, let $T$ be any linear operator in $V$ and let $J$ be its Jordan canonical form (I use the one with ones above the diagonal). Then there exists a bijective linear map $S : \mathbb C^{n\times n}\to V$ such that $T = SJS^{-1}$. I claim that $V_i = \operatorname{span}\{Se_1,Se_2,\ldots,Se_i\}$ is a chain as desired. If you know already from your lecture that the subspaces $W_i = \operatorname{span}\{e_1,\ldots,e_i\}$ form an invariant chain for $J$, then you are already done (check it!). If not, check out the following:
First, $\{V_i\}$ is obviously nested and $\dim V_i = i$ as $S$ is invertible. Concerning the invariancy, let us start with $V_1 = \operatorname{span}\{Se_1\}$. We have $TSe_1 = SJe_1 = \lambda Se_1\in V_1$, where $\lambda$ is the first eigenvalue in the Jordan form. Ok, that's settled. Now, there is either a one right to $\lambda$ in the JCF or a zero. In the second case, you have as above $TSe_2 = \mu e_2$ with $\mu$ being the second eigenvalue in the JCF (which might be $\lambda$ or not). Let us look at the first case. Then $TSe_2 = SJe_2 = S(e_1+\lambda e_2) = Se_1 + \lambda Se_2\in V_2$. So, also $V_2$ is $T$-invariant.
I hope you get the idea...
A: Let do this by induction on the dimension:


*

*In dimension 1 it is true take $\{0\}\subset V$ with $\dim V=1$

*Let suppose the result is true for very subspaces $W$ of dimension $\dim U<n$ and let prove that is true for $V$ of dimension $\dim = n$ .


Since we are in complex space $T$ has eigenvalues and can decompose as direct sum  of eigen-spaces   that is 
$$V= \bigoplus_{i =1}^{p} E_{\lambda_i}\equiv E_{\lambda_1}\oplus U$$
where we suppose that $T$ that has $p>1$ eigenvalues and let 
$$U = \bigoplus_{i =2}^{p} E_{\lambda_i}$$
$$T =T_1\oplus T'$$
Where $T_1 =T|_{E_{\lambda_1}}$ and  $T' =T|_{U}$ .

We asume $p>1:$
  We recall and it is easy to show that $T( E_{\lambda_i})\subset E_{\lambda_i}  $ since  

$$T(\ker (T-\lambda_iI))\subset \ker (T-\lambda_iI)$$
Hence One see that $$ T( E_{\lambda_1})\subset E_{\lambda_1}  $$ 
and 
$$T( U)\subset U $$ since $p>1$
we have that $$r =\dim E_{\lambda_1} <n~~~and ~~~n-r =\dim U <n$$
Whence By asumption of induction,there are two chain 
$$\color{green}{W_0⊆W_1⊆....⊆W_r=W}$$ such that $\dim W_i = i,~~i= 0,1,\cdots ,r$ and $T_1(W_i)\subset W_i$
and 
$$\color{blue}{U_0⊆U_1⊆....⊆U_{n-r}=U}$$
such that $\dim U_i = i,~~i= 0,1,\cdots ,n-r$ and $T'(U_i)\subset U_i$
Now consider the chain 
$$ \color{blue}{W_0\oplus U_0⊆W_0\oplus U_1⊆....⊆}\color{red}{W_0\oplus U_{n-r} =U\oplus W_0} \color{green}{\subset W_1\oplus U ⊆...⊆W_r\oplus U =W\oplus U=V}$$
That is $$ \begin{cases} \color{blue}{V_i~~~~~= W_0\oplus U_i}&\text{if}~~0\le i \le n-r\\
\color{green}{V_{n-r+i} = U\oplus W_i} &\text{if}~~0\le i \le r \end{cases} $$


*

*For $0\le i \le n-r$ we have
$\color{blue}{V_i =W_0\oplus U_i }$ then $\color{blue}{T(V_i) = T_1(W_0)\oplus T'(U_i) \subset W_0\oplus U_i = V_i}$
$$\color{blue}{\dim V_i = \dim U_i +\dim W_0 = i}$$

*For $0\le i \le r$ we have 
$\color{green}{V_{n-r+i} =W_i\oplus U }$ then $\color{green}{T(V_i) = T_1(W_i)\oplus T'(U) \subset W_i\oplus U = V_{n-r+i}}$


and since,$\dim U=n-r$ and $\dim W_i =i$ we  have 
$$\color{green}{\dim V_{n-r+i} = \dim U  +\dim W_{i} = n-r+i}$$
 - We have the chain 
$$V_0⊆V_1⊆....⊆V_{n}=V$$
 - $T(V_i)\subset V_i$

The cases where $p= 1$ is obvious since in that cases, it means
  $$V= \bigoplus_{i =1}^{p} E_{\lambda_i}=E_{\lambda_i} =\ker (T-\lambda_1 I)$$

i.e $$T= \lambda_1I$$
take $$V_i=\{v_1,v_2,\cdots,v_i\}$$
Where $\{v_1,v_2,\cdots,v_n\}$ is any basis of $V$
