I am given a matrix $A$ that is defined the following way:
the element in row $i$ and column $j$ is $\alpha$ if $i=j$, $1$ if $j=i+2$ and $0$ otherwise.
I need to find Jordan Form. This is my solution, I feel like this is too simple so I want to check if I understand it right:
calculating the characteristic polynomial we get that $\alpha$ is the only eigenvalue, therefore the elements in jordan form on the diagonal are all $\alpha$, so the jordan form in the $i,j$ place is $\alpha$ if $i=j$, $1$ if $i = j+1$ and otherwise $0$.
Is the solution fully correct?
No, it is not correct. One example of such a matrix is$$\begin{bmatrix}\alpha&0&1\\0&\alpha&0\\0&0&\alpha\end{bmatrix}$$and you are claiming that its Jordan normal form is$$\begin{bmatrix}\alpha&1&0\\0&\alpha&1\\0&0&\alpha\end{bmatrix},$$whereas it actually is$$\begin{bmatrix}\alpha&0&0\\0&\alpha&1\\0&0&\alpha\end{bmatrix}.$$