Let A be an $18\times 18$ matrix over $\mathbb{C}$ with characteristic polynomial equal to $$ (x-1)^6(x-2)^6(x-3)^6 $$ and a minimal polynomial equal to $$ (x-1)^4(x-2)^4(x-3)^3. $$ Assume $(A-I)$ has nullity $2$, $(A-2I)$ has nullity 3, and $(A-3I)^2$ has nullity 4. Find the Jordan canonical form of A.

For $\lambda=1$, I figure the largest block is a $4\times 4$ (since the multiplicity of the root is $4$ in the minimal polynomial), and since the nullity is $2$, that there are $2$ blocks in total, thus forcing the other block to be $2\times2$.

For $\lambda=2$, I figure the largest block is a $4\times4$, there are 3 blocks, thus making the others $1\times 1$ matrices.

For $\lambda=3$, The largest block is a $3\times 3$ matrix. However, I am not sure how the nullity of $(A-3I)^2$ determines the nullity of $(A-3I)$.

Any insight into this problem would be extremely valuable as I am trying to teach myself this process.

  • 2
    $\begingroup$ The restriction of $(A-\lambda I)^k$ to a Jordan block of size $n\times n$ (belonging to $\lambda$) has nullity $\min\{k,n\}$. So for example, if the sizes of the $\lambda=3$ blocks were $3+1+1+1=6$, then the nullity of $(A-3I)^2$ would be $$\min\{3,2\}+3\min\{3,1\}=2+3\cdot1=5,$$ which is too much. Check out the other cases. I think that only one works. $\endgroup$ – Jyrki Lahtonen Aug 17 '13 at 22:06
  • 1
    $\begingroup$ Oh, and if you have problems trusting that rule, here's how I would go about proving it: a Jordan block of size $n\times n$ has those $n-1$ ones off the diagonal. Every time you multiply it with itself (with the diagonal entries nulled), those ones move one position NorthEast, so one of the drops out. So after raising to power $k$, you have $\max\{n-k,0\}$ ones remaining. That's then also the rank of this block in $(A-\lambda I)^k$. The nullity is then $n$ minus the rank. $\endgroup$ – Jyrki Lahtonen Aug 17 '13 at 22:23
  • $\begingroup$ I would say the question I have from this arrises from the use of the formula. Are you saying the nullity is $\sum min(k,n)$ over all blocks? If so, since the multiplicity of $\lambda=3$ is 3, that guarantees a 3x3 block, correct? Then we can either have 3+3, 3+2+1, or 3+1+1+1. Using your formula, then we have min(2,3)+min(2,3), min(2,3)+min(2,2)+min(2,1), min(2,3)+3min(2,1); none of which sum to 6. Do you mean min(k,n) or do you mean something else? $\endgroup$ – user90240 Aug 17 '13 at 23:18
  • $\begingroup$ Weren't you given that the nullity of $(A-3I)^2$ is four? And $\min(3,2)+\min(3,2)=4$, so... $\endgroup$ – Jyrki Lahtonen Aug 18 '13 at 5:13
  • $\begingroup$ @JyrkiLahtonen Please consider converting your comments into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$ – Julian Kuelshammer Sep 15 '13 at 10:50

The followig auxiliary result is needed to settle this question. Let us consider an $n\times n$ Jordan block $$ B=\left(\begin{array}{ccccc} \lambda&1&0&\cdots&0\\ 0&\lambda&1&\cdots&0\\ \vdots&\ddots&\ddots&\ddots&\vdots\\ 0&\cdots&0&\lambda&1\\ 0&\cdots&\cdots&0&\lambda \end{array}\right). $$ If $0\le k\le n$, then we see that $(B-\lambda I)^k$ has a diagonal of length $n-k$ filled with $1$s. Therefore its rank is $n-k$, and thus its nullity is equal to $k$. If $k>n$, then the nullity of this block is, of course $n$. We can summarize this by saying that the nullity of $(B-\lambda I)^k$ is $\min\{n,k\}$.

The OP can use this to handle the case $\lambda=3$. Assume that the Jordan blocks associated to this eigenvalue have size $n_1\ge n_2\ge\cdots\ge n_m$. The characteristic polynomial tells us that $n_1+n_2+\cdots+n_m=6.$ For its part the minimal polynomial tells us that $n_1=3$. We are left with three possibilities: A) $m=2,n_1=n_2=3$, B) $m=3, n_1=3,n_2=2,n_3=1$, C) $m=4, n_1=3, n_2=n_3=n_4=1$. Applying the above observation to all those Jordan blocks we see that the nullity of $(A-3\lambda I)^2$ is $$ \sum_{j=1}^m\min\{n_j,2\}, $$ which adds up to $2+2=4$ in case A, $2+2+1=5$ in case B, and $2+1+1+1=5$ in case C. Thus we are left with case A as the only possibility, and can conclude that the matrix $A$ has two $3\times3$ Jordan blocks beloging to $\lambda=3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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