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This is one of the prelim questions I am having trouble with. Also I need some Jordan Canonical examples similar to the following question. Thank you an advance.

We need to find a Jordan Canonical Form for an 8x8 matrix A such that $(A-I) $ has nullity 2, $(A-I)^2 $ has nullity $4$, $ (A-I)^k $ has nullity $5$ for $ k \ge 3 $, and $(A+2I)^j $ has nullity $3$ for $ j \ge 1 $

I just know, A has $2$ Jordan blocks for eigenvalue $1$ since $\dim(\operatorname{ker}(A-I))=2$ and $3$ Jordan blocks for eigenvalue $-2$. I need help to understand this question. Thank you.

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up vote 2 down vote accepted

So far, you know that the eigenvalue $1$ has multiplicity five and the eigenvalue $-2$ has multiplicity $3$. This tells you that the total size of the Jordan blocks corresponding to $1$ is five and the total size of the Jordan blocks corresponding to $-2$ is three.

Let us look at $-2$ first. Since the nullity of $A + 2I$ is three, we know that the geometric multiplicity of $-2$ is three. Recall that the geometric multiplicity is the number of Jordan blocks corresponding to that eigenvalue. There are three blocks for $-2$ with total size three, therefore the only option we have is for all three blocks to be trivial.

Now for eigenvalue $1$, we know that the geometric multiplicity is two. Therefore there will be two Jordan blocks. The blocks must add to size five, therefore the possibilities are

  1. A block of size one and a block of size four
  2. A block of size two and a block of size three

The point at which your generalized eigenspaces stabilize (no longer increase in size) is the length of your longest chain. Your matrix stabilizes at $k = 3$ and so your longest chain is length three. Correspondingly, the size of your largest block is three and so we must default to option 2. There is a single Jordan block of size three and a Jordan block of size two corresponding to eigenvalue $1$.

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thanks a lot Eu Yu. It is very helpful. – rose Nov 15 '12 at 18:13

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