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.

  • $\begingroup$ Do you know the Jordan canonical form? $\endgroup$
    – amsmath
    Oct 15, 2017 at 13:48
  • $\begingroup$ Yes I know it.Jordan canonical form of a matrix. But how can I proceed from there? $\endgroup$
    – abcdmath
    Oct 15, 2017 at 14:00
  • $\begingroup$ I have updated my answer and it should be alright now $\endgroup$
    – Guy Fsone
    Oct 15, 2017 at 15:35

3 Answers 3


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.

  • $\begingroup$ Very nice and smart! Sometimes taking the quotient space is more helpful than just taking any complementary subspace (which here might not be $T$-invariant). $\endgroup$
    – amsmath
    Oct 15, 2017 at 16:57

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...


Let do this by induction on the dimension:

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

  • $\begingroup$ Sir what is $f$ here? $\endgroup$
    – abcdmath
    Oct 15, 2017 at 13:29
  • $\begingroup$ But how is $\ker T^n=V$ $\endgroup$
    – Learnmore
    Oct 15, 2017 at 13:38
  • $\begingroup$ Sir, do you mean that $T(Ker T)=Ker T²$. Then by your assumption $V_2⊆V_1$ which is opposite. Please elaborate it. $\endgroup$
    – abcdmath
    Oct 15, 2017 at 13:39
  • $\begingroup$ You go out from a nilpotent operator. But $T$ is any operator. Also, $\dim V_i = i$ is required. $\endgroup$
    – amsmath
    Oct 15, 2017 at 13:45
  • 2
    $\begingroup$ Nope. Try to do this with$$T = \begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}$$. Good luck! $\endgroup$
    – amsmath
    Oct 15, 2017 at 16:48

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