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 8×8 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.
Best Answer
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
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 $1$.