How to prove that A and B are similar Let be $$A=\begin{pmatrix}
\frac{-3}{2} & 2 & \frac{-1}{2} \\
\frac{-1}{2} & 0 & \frac{1}{2} \\
\frac{1}{2} & -2 & \frac{3}{2}
\end{pmatrix},
B=\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}$$
Prove that A and B are similar.
I know if we can find a matrix $P$ so that $A=P^{-1}BP$ we have that they are similar, but I haven't found anywhere how to find such $P$. Can someone give me a hint?
 A: Let $J$ be the Jordan canonical matrix that is similar to both A and B. So we have $A=S^{-1}JS$ and $B=R^{-1}JR$ 
$\Rightarrow J=RBR^{-1}$
$\Rightarrow A=S^{-1}RBR^{-1}S=P^{-1}BP$
This would be a general move you could make. 
In case you would like to check your work, your $P$ should be
\begin{bmatrix}
    1 &0&0 \\
    -\frac{3}{2}&2&-\frac{1}{2}\\
    1&-2&1
\end{bmatrix}
A: Recall that the matrix $P$ that you seek is essentially a change-of-basis matrix and that the columns of a transformation matrix are the images of the basis vectors. In this problem, $B$ has a particularly simple form, so let’s start there. The columns of $B$ tell us that, in some basis $\{v_1,v_2,v_3\}$, $v_1$ spans the kernel of the transformation, so computing $\ker A$ will give you $v_1$. The next two columns of $B$ tell us that $Bv_2=v_1$ and $Bv_3=v_2$. Thus, to find the rest of the basis, you must solve in turn $Av_2=v_1$ and $Av_3=v_2$. Once you have this basis, you can assemble the change-of-basis matrix for it.
A: The easiest way I know would be to show that $A^2$ and $B^2$ are nonzero, but $A^3=B^3=0$.  There is only one $3$-by-$3$ nilpotent matrix of nilpotency index $3$, up to conjugation.
A: Let me address this part of your post:

I know if we can find a matrix $P$ so that $A=P^{-1}BP$ we have that they are similar, but I haven't found anywhere how to find such P. Can someone give me a hint?

Notice that $A=P^{-1}BP$ is equivalent to $PA=BP$. I.e., we want an invertible matrix $P$ such that $PA=BP$.
Let us denote the rows of the matrix $P$ by $\vec r_1$, $\vec r_2$, $\vec r_3$, i.e.,
$$P=\begin{pmatrix}\vec r_1\\\vec r_2\\\vec r_3\end{pmatrix}.$$
Then we have
$$PA= \begin{pmatrix}\vec r_1A\\\vec r_2A\\\vec r_3A\end{pmatrix}
\qquad\text{and}\qquad
BP=\begin{pmatrix}\vec r_2\\\vec r_3\\\vec 0\end{pmatrix}.$$
So to get $PA=BP$ we need linearly independent vectors $\vec r_1$, $\vec r_2$, $\vec r_3$ such that
\begin{align*}
\vec r_1 A &= \vec r_2\\
\vec r_2 A &= \vec r_3\\
\vec r_3 A &= 0
\end{align*}
We can find all vectors such that $\vec r_3 A =0$ simply by solving the corresponding system of linear equations. We will get that $\vec r_3$ must be a multiple of $(1,-4,1)$. Let us choose $\vec r_3=(1,-2,1)$.
Now we have a new linear system $\vec r_2 A=\vec r_3$. One of the solutions is $\vec r_2=(0,-1,1)$.
Solving one more linear system $\vec r_1 A = \vec r_2$ we get, among other solutions, $\vec r_1=(0,1,1)$.
We can check, that for the matrix $$P=
\begin{pmatrix}
  0 & \frac12 & \frac12 \\
  0 &-1 & 1 \\
  1 &-2 & 1 \\
\end{pmatrix}
$$
we indeed have $P^{-1}AP=B$.
Here is the computation in WolframAlpha.

Above we simply tried to find $P$ directly from the condition $P^{-1}AP=B$. However, this is closely related to eigenvectors and Jordan normal form. So if you have learned some theory from this area, you will be able to check that the two matrices are similar without explicitly calculating the matrix $P$.
In fact, the last row of the matrix $P$ is row eigenvector of the matrix $A$. The first two row are generalized eigenvectors. And $B$ is Jordan normal form of the matrix $A$.
I should also mention that when I solved the three linear systems above, I chose arbitrary solution. This was basically a lucky accident. If the Jordan form were different, the same approach would not necessary work. (For example, if $A$ were similar to $B=
\begin{pmatrix}
  0 & 1 & 0 \\
  0 & 0 & 0 \\
  0 & 0 & 0 \\
\end{pmatrix}
$ and I tried to use similar approach, then I would no longer be able to choose $\vec r_2$ and $\vec r_3$ as arbitrary solutions of the linear systems I get.)
