Let $T$ be a nilpotent linear operator on the vector space $\Bbb R^5$ (i.e., $T^k = 0$ for some k).
Let $d_i$ denote the dimension of the kernel of $T^{i}$ . Which of the following can possibly occur as a value of $(d_1,d_2,d_3)$?

1. $(1,2,3)$,

2.$(2,3,5)$,

3.$(2,2,4)$,

4.$(2,4,5)$.

Let $A$ be a $5 \times 5$ nilpotent matrix. We know that its eigenvalues are $0,0,0,0,0$. By Jordan canonical form theorem, there exists an invertible matrix $P$ such that $J(A) := P A P^{-1}$ will be the Jordan canonical form of $A$. Now, $J(A)$ is a block diagonal matrix in which each block is of size varies between one to five. For example, a block of size five is of the form $$\begin{bmatrix}0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0\end{bmatrix}$$

a block of size four is of the form

$$\begin{bmatrix}0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \\ 0&0&0&0\end{bmatrix}$$

and so on for the blocks of size $3,2,1$.

Proof: $B$ is a $k \times k$ matrix with first $k-1$ rows are linearly independent and the last row is zero. The result follows.

$$\begin{bmatrix}0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0\end{bmatrix}$$

Then $B^2, B^3,B^4$ and $B^5$ are given respectively by the following matrices

$$\begin{bmatrix}0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

$$\begin{bmatrix}0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

$$\begin{bmatrix}0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

$$\begin{bmatrix}0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

Note that the above matrices are not in their Jordan canonical form. So power of a Jordan block need not be in the Jordan form again. We observe that the Jordan block $B$ has one zero row and when we take the powers every step the last non-zero row is converted into a new zero row. Hence for $1 \le k \le 5$, $$N(B^k) = k.$$

Proof: Let $J(A)$ equal to

$\begin{bmatrix}B_1&0&0&0&0 \\ 0&B_2&0&0&0 \\ 0&0&B_3&1&0 \\ 0&0&0&B_4&0 \\ 0&0&0&0&B_5\end{bmatrix}$

be the Jordan canonical form of the $n \times n$ matrix $A$. We have $5$ blocks in this decomposition. Let their sizes be $k_1,k_2,k_3,k_4$ and $k_5$ then $k_1 + \cdots +k_5 = n$. In particular the sizes of the blocks form a partition of $n$. For example, if we consider a $5 \times 5$ nilpotent matrix then the possible Jordan block sizes in the Jordan canonical form are $5$

(one single $5$ block), $4+1$ (one $4$ block and one $1$ block ), $3+2$, $3+1+1$, $2+2+1$, $2+1+1+1$, $1+1+1+1+1$ given by all the partitions of $5$. Now, by the observation 1, $N(B_i) = 1$ for all $1 \le i \le 5$. Since $J(A)$ is a block diagonal matrix with blocks $B_1,\dots,B_5$ we have $N(J(A)) = N(B_1)+N(B_1)+\cdots+N(B_5) = 5$. Since $A$ and $J(A)$ are similar matrices we have $N(A) = N(J(A)) = 5 = \text{ number of blocks in }J(A)$. This proves the observation.

$$\begin{bmatrix}B_1&0&0&0&0 \\ 0&B_2&0&0&0 \\ 0&0&B_3&1&0 \\ 0&0&0&B_4&0 \\ 0&0&0&0&B_5\end{bmatrix}^k $$ is equal to

$$\begin{bmatrix}B_1^k&0&0&0&0 \\ 0&B_2^k&0&0&0 \\ 0&0&B_3^k&1&0 \\ 0&0&0&B_4^k&0 \\ 0&0&0&0&B_5^k\end{bmatrix}.$$

With these observations we will investigate the options.

$$\begin{bmatrix}0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0\end{bmatrix}$$

Then $A^2, A^3$ are given respectively by the matrices

$$\begin{bmatrix}0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

$$\begin{bmatrix}0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

Clearly $N(A^2) = 2$ and $N(A^3) = 3$.

Suppose $J(A)$ has block decomposition $4+1$. Let $B_1$ be the size $4$ block and $B_2$ be the size $1$ block in $J(A)$. When we find $(J(A))^2$, the last non-zero row of $B_1$ will be changed to zero-row. Since $B_2$ is a $1 \times 1$ block it won't have a non-zero row to change. Hence $N((J(A)^2) = 2 + 1 = 3$. Again, when we find $(J(A))^3$, similar argument can be appliead and $N((J(A)^3) =3 + 1 = 4$. But we want this number to be $5$. Hence the block size $4+1$ is also not possible for option b. Since $3+2$ and $4+1$ are the only two possible block sizes we have proves that

A pictorial approach of the same can be seen here:

https://maksmaths.blogspot.com/2020/04/nbhm-problem.html

Click 1. $(1,2,3)$,

2.$(2,3,5)$,

3.$(2,2,4)$,

4.$(2,4,5)$.

**I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.****Solution:**Let $A$ be a $5 \times 5$ nilpotent matrix. We know that its eigenvalues are $0,0,0,0,0$. By Jordan canonical form theorem, there exists an invertible matrix $P$ such that $J(A) := P A P^{-1}$ will be the Jordan canonical form of $A$. Now, $J(A)$ is a block diagonal matrix in which each block is of size varies between one to five. For example, a block of size five is of the form $$\begin{bmatrix}0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0\end{bmatrix}$$

a block of size four is of the form

$$\begin{bmatrix}0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \\ 0&0&0&0\end{bmatrix}$$

and so on for the blocks of size $3,2,1$.

**Observation1**: If $B$ is a size $k$ Jordan block of $A$ then $\text{N}(B) := \text{ Nullity of $B$} = \text{dimension of Kernal of $B$}= 1$ and $\text{R}(B) := \text{rank of $B$} = \text{dimension of Range of $B$} = k-1$.Proof: $B$ is a $k \times k$ matrix with first $k-1$ rows are linearly independent and the last row is zero. The result follows.

**Observation2**: Let $B$ be the following size $5$ Jordan block.$$\begin{bmatrix}0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0\end{bmatrix}$$

Then $B^2, B^3,B^4$ and $B^5$ are given respectively by the following matrices

$$\begin{bmatrix}0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

$$\begin{bmatrix}0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

$$\begin{bmatrix}0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

$$\begin{bmatrix}0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

Note that the above matrices are not in their Jordan canonical form. So power of a Jordan block need not be in the Jordan form again. We observe that the Jordan block $B$ has one zero row and when we take the powers every step the last non-zero row is converted into a new zero row. Hence for $1 \le k \le 5$, $$N(B^k) = k.$$

**Observation3**: Let $A$ be a $n \times n$ nilpotent matrix, then the number of Jordan blocks in the Jordan decomposition of $A$ is equal to $N(A)$ the dimension of the kernal of $A$.Proof: Let $J(A)$ equal to

$\begin{bmatrix}B_1&0&0&0&0 \\ 0&B_2&0&0&0 \\ 0&0&B_3&1&0 \\ 0&0&0&B_4&0 \\ 0&0&0&0&B_5\end{bmatrix}$

be the Jordan canonical form of the $n \times n$ matrix $A$. We have $5$ blocks in this decomposition. Let their sizes be $k_1,k_2,k_3,k_4$ and $k_5$ then $k_1 + \cdots +k_5 = n$. In particular the sizes of the blocks form a partition of $n$. For example, if we consider a $5 \times 5$ nilpotent matrix then the possible Jordan block sizes in the Jordan canonical form are $5$

(one single $5$ block), $4+1$ (one $4$ block and one $1$ block ), $3+2$, $3+1+1$, $2+2+1$, $2+1+1+1$, $1+1+1+1+1$ given by all the partitions of $5$. Now, by the observation 1, $N(B_i) = 1$ for all $1 \le i \le 5$. Since $J(A)$ is a block diagonal matrix with blocks $B_1,\dots,B_5$ we have $N(J(A)) = N(B_1)+N(B_1)+\cdots+N(B_5) = 5$. Since $A$ and $J(A)$ are similar matrices we have $N(A) = N(J(A)) = 5 = \text{ number of blocks in }J(A)$. This proves the observation.

**Observation4**: The $k$th power of $J(A)$$$\begin{bmatrix}B_1&0&0&0&0 \\ 0&B_2&0&0&0 \\ 0&0&B_3&1&0 \\ 0&0&0&B_4&0 \\ 0&0&0&0&B_5\end{bmatrix}^k $$ is equal to

$$\begin{bmatrix}B_1^k&0&0&0&0 \\ 0&B_2^k&0&0&0 \\ 0&0&B_3^k&1&0 \\ 0&0&0&B_4^k&0 \\ 0&0&0&0&B_5^k\end{bmatrix}.$$

With these observations we will investigate the options.

**option 1**:(1,2,3)(**Possible**) It is given that $d_1 = 1$, $d_2 = 2$ and $d_3 = 3$. By the observation 2, we have to find a $5 \times 5$ matrix $A$ such that $J(A)$ has one Jordan block, $N(A^2) = 2$ and $N(A^3) = 3$. Let $A$ be the $5\times 5$ matrix given below which itself is in the Jordan canonical form.$$\begin{bmatrix}0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0\end{bmatrix}$$

Then $A^2, A^3$ are given respectively by the matrices

$$\begin{bmatrix}0&0&1&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

$$\begin{bmatrix}0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0\end{bmatrix}$$

Clearly $N(A^2) = 2$ and $N(A^3) = 3$.

**option 2**:**(2,3,5)**(**Not Possible**) It is given that $d_1 = 2$, $d_2 = 3$ and $d_3 = 5$. By the observation 2, we have to find a $5 \times 5$ matrix $A$ such that $J(A)$ has two Jordan blocks, $N(A^2) = 3$ and $N(A^3) = 5$. Since $J(A)$ has two Jordan blocks, the possible block sizes are $3+2$ or $4+1$. Suppose $J(A)$ has block decomposition $3+2$. Let $B_1$ be the size $3$ block and $B_2$ be the size $2$ block in $J(A)$. When we find $(J(A))^2$, the last non-zero row of $B_1$ will be changed to zero row and similarly in $B_2$. Hence $N((J(A)^2) = N(J(A))+2 = 2+2 = 4$. But we want this number to be $3$ for option b. Hence the block size $3+2$ is not possible for option b. We continue to calculate $N((J(A))^3)$ which will be helpful for option d. When we find $(J(A))^3$, again the last non-zero row will be changed to zero row in $B_1$. Since $B_2$ is a $1 \times 1$ block it wont have a non-zero row to change. Hence $N((J(A)^3) = N((J(A))^2)+1 = 4+1 = 5$.**Hence $(d_1,d_2,d_3) = (2,4,5)$ is possible which is option d.**Suppose $J(A)$ has block decomposition $4+1$. Let $B_1$ be the size $4$ block and $B_2$ be the size $1$ block in $J(A)$. When we find $(J(A))^2$, the last non-zero row of $B_1$ will be changed to zero-row. Since $B_2$ is a $1 \times 1$ block it won't have a non-zero row to change. Hence $N((J(A)^2) = 2 + 1 = 3$. Again, when we find $(J(A))^3$, similar argument can be appliead and $N((J(A)^3) =3 + 1 = 4$. But we want this number to be $5$. Hence the block size $4+1$ is also not possible for option b. Since $3+2$ and $4+1$ are the only two possible block sizes we have proves that

**$(d_1,d_2,d_3) = (2,3,5)$ is not possible**.**option c:(2,2,4)(Not Posssible)**The argument given in option $2$ shows that if $d_1 = 1$ then the possible $(d_1,d_2,d_3)$ are $(2,3,4)$ and $(2,4,5)$. Hence $(2,2,4)$ is not possible.**option d:(2,4,5)(Possible)**This is done as part of option b.A pictorial approach of the same can be seen here:

https://maksmaths.blogspot.com/2020/04/nbhm-problem.html

**here**for more problems.

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Excellent proof prof. I have done a proof on the fact "A linear transformation is completely determined by its behaviour on a basis.".

ReplyDeletehttps://maksmaths.blogspot.com/2020/04/nbhm-problem.html

Thanks and Very good. Yours is more pictorial and of course conceptual. Let me tag it in the blogs that I follow.

ReplyDeleteI am still a student and your writing shows more maturity. So you must be prof and call me friend.

DeleteAny difficulty with reading the blog? if so pls tell me. I did changes.

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