I'm currently working on a linear algebra problem concerning Jordan forms. Indeed I'm given the matrix:
$$A = \left(\begin{array}{ccccccc}
1 & 0 & 1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 2 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 2 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 2 & 0 & 1\\
0 & 0 & 0 & 0 & 0 & 2 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 2\\
\end{array}\right)$$
For which I must find its Jordan form. I'm stuck at the end actually, so let me tell you what i found so far:
The characteristic polynomial of $A$ is just $\chi_A(\lambda) = (\lambda – 1)^2(\lambda – 2)^5$. So denote$\lambda_1 = 1$ of algebraic multiplicity $1$ and $\lambda_2 = 2$ of algebraic multiplicity $5$.
I then determined the eigenspaces $E_1 = Span\{(1,0,0,0,0,0,0);(0,1,0,0,0,0,0)\}$ and $E_2 = Span\{(1,0,1,0,0,0,0);(0,1,0,1,0,0,0)\}$ associated to $\lambda_1$ and $\lambda_2$, respectively, which both have dimension 2. And since this dimension corresponds to the geometric multiplicity of the associated eigenvalue, we observe that algebraic and geometric multiplicities of $\lambda_1$ coincides, thus the Jordan block with eigenvalue $\lambda_1$ on the diagonal will be composed of two subblocks of dimension $1$, i.e: $\left(\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right)$.
However algebraic and geometric multiplicities of $\lambda_2$ do not coincide ($5 \ne 2$). I've seen that to find how to represent the Jordan block in this case, we must find the dimension of $\ker((A – \lambda_2I_7)^2)$ and if it's 5, we stop, but if not need to determine dimension of $\ker((A – \lambda_2I_7)^3)$ and so on until we reach 5.
I think I know how to determine the size of Jordan subblocks from there, but my problem is that I always find $dim(\ker((A – \lambda_2I_7)^n))$ to be odd… so it skips $5$, and I don't know if there is a mistake in my argument or in my computation, could anyone take a look at this and tell me? It would reaaly help me!
Best Answer
Indeed, you seem to be making a mistake. You should find that $$ \dim \ker(A - 2 I) = 2, \quad \dim \ker(A - 2 I)^2 = 4, \\ \dim \ker(A - 2 I)^n = 5 \qquad n \geq 3. $$ Correspondingly, you should find that there is a block of size $2$ and a block of size $3$ associated with the eigenvalue $2$.