NBHM 2020 PART C Question 26 Solution (Jordan canonical form and Nilpotent matrices)(NBHM interview concept)

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.$(2,4,5)$.

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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

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4 comments:

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

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

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

    ReplyDelete
    Replies
    1. I am still a student and your writing shows more maturity. So you must be prof and call me friend.

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

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