# Find the rank of $T^2$

Question: Let $$\mathbb{C}^{11}$$ is a vector space over $$\mathbb{C}$$ and $$T:\mathbb{C}^{11}\to \mathbb{C}^{11}$$ is a linear transformation. If dimension of Kernel $$T=4$$, dimension of Kernel $$T^3=9$$ and dimension of Kernel $$T^4=11$$. Then the dimension of Kernel $$T^2=$$............

Since $$T$$ is a linear operator, $$T^2, T^3,T^4$$ will also be linear operators and there will be matrices associated with these linear operators, say $$[T]$$ represents the matrix related to the linear operator $$T$$. By rank-nullity theorem, we get $$rank(T)+nullity(T)=dim(\mathbb{C}^{11})=11$$. So, rank$$(T)=7$$ and similarly, we can get rank$$(T^3)=2$$ and rank$$(T^4)=0$$. Therefore, $$T$$ is nilpotent. Again by rank-nullity theorem, $$nullity(T^2)=dim(\mathbb{C}^{11})-rank(T^2)=11-rank(T^2)$$. Now the main problem is reduced to find the rank of $$T^2$$.

We know that $$T$$ is nilpotent. Now, let $$B_{11 \times 11}=T^2$$ and $$B^2=T^4=0$$, then the rank of $$B$$ can be found using this fact Matrix algebra: If $A^2=0$, Proof rank(A) $\le \frac{n}{2}$ . We get, rank$$(T^2)\leq \frac{11}{2}$$. In this way we can tell possibilities of the dimension of Kernel of $$T^2$$.

Can we only find the possibilities not the exact rank of $$T^2$$ with the given data?

• Do you know about the Jordan form of nilpotent matrices? May 13 '20 at 11:18
• There is a fairly simple argument, but before going into details I would like to know is this is a home work assignment? May 15 '20 at 20:24
• Quite the problem. Thanks for posting this! May 15 '20 at 21:43

Let $$Z_k={\rm ker} T^k$$, $$k\geq 0$$ be the sequence of kernel spaces. One has $$Z_0 =\{0\} \subset Z_1 \subset Z_2 ...$$ Let $$d_k = {\rm dim\ } Z_k$$. Then one has the following inequality for $$k\geq 1$$: $$d_{k+1}-d_k \leq d_k-d_{k-1}$$ In your case this yields $$11-9\leq 9-d_2\;$$ or $$\;d_2\leq 7$$ and $$\;9-d_2\leq d_2-4\;$$ or $$\;d_2\geq 6.5$$. The unique integer solution is thus $$d_2=7$$ and by the kernel-rank theorem we get $${\rm rank}\; T^2=4$$ in accordance with the statement of Alex.

I imagine the above inequality is well-known by specialists but I don't have a reference for it. To prove it note that since $$T Z_{k+1}\subset Z_k$$ and $$T Z_k\subset Z_{k-1}$$ we have a well-defined map between quotients: $$\widehat{T} : Z_{k+1}/Z_k \rightarrow Z_k/Z_{k-1} .$$ I claim this map is injective. If $$w\in Z_{k+1} \setminus Z_k = Z_{k+1} \setminus T^{-1}(Z_{k-1})$$ then the last expression shows indeed that $$Tw\in Z_k\setminus Z_{k-1}$$. As the map is injective dimensions must increase so $$d_{k+1}-d_k = {\dim\;} Z_{k+1}/Z_k \leq \dim Z_k/Z_{k-1}=d_k-d_{k-1}.$$ If you don't fancy quotient spaces you may concoct a (slightly longer) proof using complements, i.e. writing $$Z_{k+1} = Z_k \oplus W_k$$ and look at how $$T$$ acts upon $$W_k$$.

About generality: The above inequality (showing concavity of $$k\mapsto d_k$$) holds in a space of any dimension (also infinite, as long as $$d_1$$ is finite). Thus the conclusion for $$d_2$$ is independent of the dimension of the ambient space. But the conclusion for the rank of $$T^2$$ is obviously not.

• This is an elegant answer. Isn't it supposed to be $d_k=\text{ dim} Z_k$ or in other words $d_k= \text{dim Ker} T^k$?? May 16 '20 at 9:13
• Indeed a typo. Thanks! May 16 '20 at 11:19
• Could you please suggest readings(book or notes) for this? Aug 4 at 6:18
• @Learning Unfortunately I don't have any. At some point I just wondered how it works and came up with the above argument (useful when Jordan normalizing a matrix). Aug 4 at 19:35

Following levap’s guide consider a $$11\times 11$$ complex matrix $$T$$ such that $$\operatorname{rank} T=7$$, $$\operatorname{rank} T^3=2$$, and $$\operatorname{rank} T^4=0$$. Let Jordan form of the matrix $$T$$ contains $$a_i$$ Jordan cells of size $$i$$ for each $$1\le i\le 11$$. Then we have the following system of equations.

$$\begin{cases} \sum_{i=1}^{11} ia_i =11\\ \sum_{i=2}^{11} (i-1)a_i =7\\ \sum_{i=4}^{11} (i-3)a_i =2\\ \sum_{i=5}^{11} (i-4)a_i =0 \end{cases}$$

It follows $$a_i=0$$ for $$i\ge 5$$ and

$$\begin{cases} a_1+2a_2+3a_3+4a_4=11\\ a_2+2a_3+3a_4 =7\\ a_4=2\end{cases}$$

We find consecutively from this system $$a_4=2$$, $$a_3=0$$, $$a_2=1$$, and $$a_1=1$$. So $$\operatorname{rank} T^2=\sum_{i=3}^{11} (i-2)a_i =2a_4=4.$$

In general, $$\ker T \subseteq \ker T^2 \subseteq \ker T^3 \subseteq \cdots$$. This implies that $$4 \leq \text{nullity}(T^2) \leq 9$$. Also in general, if $$T^n = T^{n + 1}$$ for some nonnegative integer $$n$$, then $$\ker T^n = \ker T^{n + 1} = \ker T^{n + 2} = \cdots$$. Hence, $$\text{nullity}(T^2)$$ cannot be 4 or 9, so $$5 \leq \text{nullity}(T^2) \leq 8$$. Since $$(T^2)^2 = T^4 = 0$$, the range of $$T^2$$ is contained in its kernel, so $$\text{nullity}(T^2) \geq \text{rank}(T^2)$$. By the rank-nullity theorem, $$11 = \text{rank}(T^2) + \text{nullity}(T^2) \geq 2 \, \text{rank}{(T^2)} \implies \text{rank}(T^2) \leq 11/2 = 5.5$$. Hence $$\text{rank}(T^2) \leq 5$$. So $$\text{nullity}(T^2) = 11 - \text{rank}(T^2) \geq 11 - 5 = 6$$. So we can say the possibilities for $$\text{nullity}(T^2)$$ are 6, 7 or 8.

• $5 \leq \text{nullity}(T^2) \leq 8$. The possibilities for the dimension of Kernel $T^2=5,6,7,8$. Let $\text{nullity}(T^2)=7 \implies rank (T^2)=4$, which contradicts your answer. May 13 '20 at 11:51
• @Learning You're right, I made a silly mistake, sorry about that. I've edited it. May 13 '20 at 12:18