Jordan normal Form and Transformation matrix

Question

I am solving an example to find the Jordan form and the transformation matrix of the following $4 \times 4$ matrix:

\begin{pmatrix}a&0&1&0\\ 1&a&1&1\\ 0&0&a&0\\ 0&0&1&a\end{pmatrix} I went through the solution manually as follows:

(1). Eigenvalue $(a- \lambda)^4$, so $\lambda=a$ with an algebraic multiplicity of $4$.

(2). Geometric multiplicities

$$\ker(A-\lambda I) = \begin{pmatrix}1&0&1&1\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix} $$

so $\dim(\ker(A-\lambda I)) = 2$ and

$$\ker(A- \lambda I) = \text{span}\{\begin{pmatrix}0\\ 1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}-1\\ 0\\ 0\\ 1\end{pmatrix}\}$$

$$ \ker(a-\lambda I)^2 = \begin{pmatrix}0&0&2&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix} $$

so $\dim(\ker(A-\lambda I)^2) = 3$ and $$\ker(A-\lambda I)^2 = \text{span}\{\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 0\\ 0\\ 1\end{pmatrix}\}$$

$$ \ker(a-\lambda I)^3 = \begin{pmatrix}0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix} $$

so $\dim(\ker(A-\lambda I)^3) = 4$ and $$\ker(A-\lambda I)^3 = \text{span}\{\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}0\\ 0\\ 0\\ 1\end{pmatrix}\}$$

(3). Transformation matrix $w_3 \in \ker(A-\lambda I)^3$, and $w_3 \not\in \ker(A-\lambda I)^2$

So let we say,

$w_{k-1}=(A-\lambda I) \cdot w_k$

so I can get:

$$ w3=\begin{pmatrix}0\\ 0\\ 0\\ 1\end{pmatrix}, w2=\begin{pmatrix}0\\ 1\\ 0\\ 0\end{pmatrix}, w2=\begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix} and u1 = \begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix}$$

So the Jordan form is $$\begin{pmatrix}a&1&0&0\\ 0&a&1&0\\ 0&0&a&0\\ 0&0&0&a\end{pmatrix}$$

the Trans. matrix is $$\begin{pmatrix}0&0&0&1\\ 0&1&0&0\\ 0&0&0&0\\ 1&0&0&0\end{pmatrix}$$

I am not sure of my hand solution, but using the Matlab function, gives

V =

                         [0    1    -2    -1]
                         [                  ]
                         [2    0     0     0]
                         [                  ]
                         [0    0     1     0]
                         [                  ]
                         [0    1     1     1]

J =

                          [a    1    0    0]
                          [                ]
                          [0    a    1    0]
                          [                ]
                          [0    0    a    0]
                          [                ]
                          [0    0    0    a]

I want to know what is the wrong step in my hand solution to get the same result! Thank you!