Given matrices $\mathbb{A} = \begin{pmatrix} 0 & 1 & \alpha\\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, \alpha \in \mathbb{C}$ and $\mathbb{B} = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ find if they are similar.
First, characteristic polynomial of both matrices is $-x^3$ and their spectrum is just $\{0\}$. To show that that they are similar I have to find regular matrix $\mathbb{X}$ such that $\mathbb{X}^{-1}\mathbb{A}\mathbb{X} = \mathbb{B}$.
Here is where I'm stuck. I don't know what to do next. When I'm diagonalising matrix, diagonal matrix consists of eigenvalues on diagonal, and similarity transform matrix is made of eigenvectors corresponding to eigenvalues. That holds only when geometric multiplicity of eigenvalues is identical to algebraic multiplicity of eigenvalue. In this case, the eigenvector is $\mathbb{\nu} = (1, 0, 0)$, so geometric multiplicity is 1, whilst algebraic is 3. But, according to my textbook solutions they are similar.
How can I prove similarity?
It is not strictly necessary to find a similarity $X$ such that $B = X^{-1}AX$ if you have already established Jordan form. In this case, it would suffice to note that $A$ has $0$ as its only eigenvector with a geometric multiplicity of $1$. From there, the only $3 \times 3$ Jordan form matrix with those properties is the matrix $B$.
If we did want to find an explicit similarity, however, we would do it by exploiting chains of generalized eigenvectors: note that $B$ is such that $$ B \pmatrix{0\\0\\1} = \pmatrix{0\\1\\0} \qquad B \pmatrix{0\\1\\0} = \pmatrix{1\\0\\0} \qquad B \pmatrix{1\\0\\0} = 0 $$ Similarly, we have $$ A \pmatrix{0\\1\\0} = \pmatrix{1\\0\\0} \qquad A \pmatrix{1\\0\\0} = 0 $$ and to obtain a suitable basis, we need a vector $v$ such that $$ Av = \pmatrix{0\\1\\0} $$ Solve this equation to find (among the may possible solutions) $$ v = \pmatrix{0\\-\alpha\\1} $$ All together, we can say that $$ X = \pmatrix{1&0&0\\0&1&-\alpha\\0&0&1} $$ is the required similarity matrix.