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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?

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    $\begingroup$ Do you know about the Jordan form of nilpotent matrices? $\endgroup$
    – levap
    May 13, 2020 at 11:18
  • $\begingroup$ There is a fairly simple argument, but before going into details I would like to know is this is a home work assignment? $\endgroup$
    – H. H. Rugh
    May 15, 2020 at 20:24
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    $\begingroup$ Quite the problem. Thanks for posting this! $\endgroup$ May 15, 2020 at 21:43

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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.

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  • $\begingroup$ This is an elegant answer. Isn't it supposed to be $d_k=\text{ dim} Z_k$ or in other words $d_k= \text{dim Ker} T^k$?? $\endgroup$
    – Learning
    May 16, 2020 at 9:13
  • $\begingroup$ Indeed a typo. Thanks! $\endgroup$
    – H. H. Rugh
    May 16, 2020 at 11:19
  • $\begingroup$ Could you please suggest readings(book or notes) for this? $\endgroup$
    – Learning
    Aug 4, 2021 at 6:18
  • $\begingroup$ @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). $\endgroup$
    – H. H. Rugh
    Aug 4, 2021 at 19:35
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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 $$\operatorname{rank} T^2=\sum_{i=3}^{11} (i-2)a_i =2a_4=4.$$

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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.

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  • $\begingroup$ $5 \leq \text{nullity}(T^2) \leq 8$. 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. $\endgroup$
    – Learning
    May 13, 2020 at 11:51
  • $\begingroup$ @Learning You're right, I made a silly mistake, sorry about that. I've edited it. $\endgroup$
    – twosigma
    May 13, 2020 at 12:18

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