Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be a $7\times 7$ matrix such that $ (A-I)^3=0$ and $(A-I)^2 $ has rank $2$. How can we find the Jordan normal form of $A$?

share|improve this question

2 Answers 2

up vote 4 down vote accepted

Since $(A-I)^3=0$, The only eigenvalue of $A$ is 1, and the Jordan blocks are of size no bigger than 3. Since $(A-I)^2$ has rank 2, There must be some Jordan block of rank bigger than 2. But if $B$ is a jordan block of size 3, $(B-I)^2$ has rank one. Hence we must have 2 Jordan blocks of size 3. Since $A$ is $7\times 7$, there is one additional Jordan block, of size 1. $$\begin{pmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 & 0 & 0\\0 & 0 & 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$

If you don't see how I arrived at this solution, try squaring and cubing the above matrix to see what happens,and it should become more clear.

share|improve this answer
    
Thanks Frankel and Jordan. By using your answer I try to solve this question. A is 4x4 matrix with complex coefficient such that $ (A-3I)^2=0 $ possible jordan forms of A. The only eigenvalue 3 and the size of Jordan blocks are not bigger than 2 since $(A-3I)^2=0. $ $ \begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1\\ 0 & 0 & 0 & 3 & \end{pmatrix} $ $ \begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 3 & \end{pmatrix} $ $ \begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 1& 0 \\ 0 & 0 & 3 & 1\\ 0 & 0 & 0 & 3 & \end{pmatrix} $ Is it right? –  the code Feb 25 '12 at 3:53
    
What you said, in words, is correct. Of the matricies you wrote, only the first is a possible solution. The second has a Jordan block of size 3, and the third is a Jordan block of size 4. There are other solutions: you could 4 Jordan blocks of size 1 (ie a diagonal matrix), or one of size 2 and two of size 1. –  Brett Frankel Feb 25 '12 at 4:01
    
$ \begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1\\ 0 & 0 & 0 & 3 & \end{pmatrix} $ $ \begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 3 & \end{pmatrix} $ $ \begin{pmatrix} 3 & 0 & 0 & 0 \\ 0 & 3 & 0& 0 \\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 3 & \end{pmatrix} $ –  the code Feb 25 '12 at 4:49
    
Thank you so much Frankel I think I get it. Now it is completed right. –  the code Feb 25 '12 at 4:51
    
Yes, this is correct. There are two other solutions, but they are essentially the same as your second solution, just with the Jordan blocks in different orders. –  Brett Frankel Feb 25 '12 at 4:58

All of $A$'s eignevalues are $1$, since $A$'s minimal polynomial divides $(X-1)^3$. So the Jordan blocks have $1$s down the diagonal. The blocks most be less than or equal to $3$ in size, or else $(A-I)^3$ would not be $0$. We could not have all of the blocks being of size $\leq2$, or else $(A-I)^2$ would already be $0$ and have $0$ rank. And each block of size $3$ can only contribute $1$ to the rank of $(A-I)^2$. So we must have two blocks of size $3$ and one of size $1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.