# Finding the Jordan form of a $3\times 3$ matrix

I'm confused: For a matrix with one repeated eigenvalue say $\lambda$, the jordan block for this matrix will look like depending on the nullities of $(A-\lambda I)^n$, doesn't it? I'll give an example:
$$A = \begin{pmatrix} 4 & -2 & 6 \\ -2 & 5 & 1 \\ -1 & 1 & 3 \end{pmatrix}$$

Its eigenvalues are $\lambda = 4,4,4$.

I found that the nullity of $(A-4I)$ is 1. This tells us we have one Jordan Block yes? So that means it must be $$J = \begin{pmatrix} 4 & 1 & 0 \\ 0 & 4 & 1 \\ 0 &0 &4\end{pmatrix}$$

But, when I compute the nullity of $(A-4I)^2$, this equals to $2$. I note the difference is $1$. Isn't this telling me that there is one Jordan block with size $2$ or greater? So this says that we only have ONE Jordan Block with size 1... which isn't true.

So basically my confusion is: I calculated the difference between nullities of $A-4I$ and $(A-4I)^2$, and this equals 1. Isn't this telling me that there is one jordan block of size 1?

• Look at $J-4I$; clearly nullity 1 and one 3x3 Jordan block; square it, clearly nullity 2 and two Jordan blocks, one size 1x1 and one size 2x2; cube it, clearly nullity 3 and three Jordan blocks, each size 1x1. So from "Isn't this telling me" you're astray. – ancientmathematician May 22 '17 at 6:42

Using the theorem: Let $T : V \rightarrow V$ be a linear map admitting a JFC, $J$. Let $i < 0$ and $\lambda \in \mathbb{C}$. Then the number of Jordan blocks with eigenvalue $\lambda$ and length at least $i$ is equal to nullity$((T-\lambda I_n)^{i}) -$nullity$((T-\lambda I_n)^{i-1})$. So, in your case, the number of $J$-Blocks with $\lambda =4$ and length $2$ is nullity$((T-\lambda I_n)^{2}) -$nullity$((T-\lambda I_n)^{1}) = 2-1 =1$. So there is one block of at least length $2$. However, nullity$((T-\lambda I_n)^{3}) -$nullity$((T-\lambda I_n)^{2})= 3-2=1$. So there is one block of length 3, implying that the Jordan normal form is $$J = \begin{pmatrix} 4 & 1 & 0 \\ 0 & 4 & 1 \\ 0 &0 &4\end{pmatrix}$$
The fact that nullity$(A-4I)=1$ does not imply that there is one Jordan block.