I am trying to understand an example of computation of Jordan Normal Form from Jim Hefferon's book "Linear Algebra". The example is given here. I am a bit confused trying to understand the similarity diagram from the example:
$\qquad\qquad\qquad $
In the end, he says the computations needed to compute the JNF are: $$\left( \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{4} & \frac{1}{4} \\ \end{array} \right) \left( \begin{array}{cc} 2 & -1 \\ 1 & 4 \\ \end{array} \right) \left( \begin{array}{cc} 1 & -2 \\ 1 & 2 \\ \end{array} \right)$$
I understand the first and third matrices, but why is it $\left( \begin{array}{cc} 2 & -1 \\ 1 & 4 \\ \end{array} \right)$ and not $\left( \begin{array}{cc} -1 & -1 \\ 1 & 1 \\ \end{array} \right)$? In the similarity diagram, the upper arrow has $T-3I$written. And $T-3I=\left( \begin{array}{cc} -1 & -1 \\ 1 & 1 \\ \end{array} \right)$.
Although, computing the following product of matrices, I get:
$$\left( \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{4} & \frac{1}{4} \\ \end{array} \right) \left( \begin{array}{cc} -1 & -1 \\ 1 & 1 \\ \end{array} \right) \left( \begin{array}{cc} 1 & -2 \\ 1 & 2 \\ \end{array} \right)=\left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array} \right)$$
Which is a matrix he mentions previously:
$$\mathtt{Rep}_{B,B}(t-3)=N=\left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array} \right)$$
So I may be doing some silly mistake but am unable to spot it. If someone need the book too, here it is, this example is on page 449 in the book or page 459 in the PDF.
You want to find a change of basis matrix $R$ such that $R^{-1} T R$ is in the JNF. The algorithm to find the basis vectors is nicely described in the case of the eigenvalue $0$, so you are really trying to find $R$ such that $R^{-1} (T-3I) R$ is in JCF. But once you find such $R$, you can just apply it to $T$ instead of $T-3I$, because by distributivity of matrix multiplication, $R^{-1} (T-3I) R = R^{-1} T R - 3R^{-1} I R = R^{-1} T R - 3 I$, and so if $R^{-1} (T-3I) R$ is in JNF, so is $R^{-1} T R = R^{-1} (T-3I) R + 3I$.
This also corresponds with your observation that for your matrix choice you got the Jordan cell with zeros on the diagonal, but when you start with the original $T$, you get a Jordan cell with threes on the diagonal.