Trouble finding Jordan Normal form for $4 \times $ 4 matrix

Question

$M = \left(\begin{array}{cccc}0 & 1 & 0 & 0 \\-3 & 4 & 0 & 0 \\2 & -1 & 2 & 0 \\-1 & 1 & 1 & 2\end{array}\right)$.

I find the eigenvalues to be $\lambda = 2,2,3,1$.

Let $d_{i,\lambda}$ be the nullity of $(M - \lambda I)^i$.

I find that $d_{1,2} = 1, d_{2,2} = 2, d_{1,3} = 1, d_{2,3} = 1, d_{1,1} = 1, d_{2,1}=1$.

This can't be right. How is this supposed to be written in Jordan normal form?

Answer 1

After computing the characteristic polynomial, and having gone through all the kernels of the generalized eigenspaces you can use the following formula to get the JNF directly:

The number of Jordan Blocks for the eigenvalue $\lambda$ of size $k\times k$ is equal to: $$2\dim\ker(A-\lambda)^k-\dim\ker(A-\lambda)^{k+1}-\dim\ker(A-\lambda)^{k-1}$$

In your case, the computation is fairly easy, indeed you only have to use the fact that:

$$\dim\ker(A-\lambda)=\dim E_\lambda=\# \text{ Jordan Blocks for the Eigenvalue } \lambda$$

And since you only have one Eigenvalue having multiplicity greater than 1, you just need to check if its Eigenspace has dimension $1$ or $2$. In the second case your matrix will be simply diagonalizable, in the first, you will have a Jordanblock of size $2$ ($\begin{pmatrix}\lambda & 1 \\ 0 &\lambda \end{pmatrix}$)

Answer 2

Based on the eigenvalues, for the Jordan normal form there are only two possibilities now: $$\pmatrix{1 &&&\\&3&&\\&&2&\\&&&2}\ \text{ or }\ \pmatrix{1 &&&\\&3&&\\&&2&0\\&&1&2} $$.

To find out, which one will be yours, all you have to do is to determine the nullity of $A-2I$.