<s>[INST] How to show that two matrices are similar without using anything about eigenvalues or eigenvectors Problem: Show that $A$ and $B$ are similar without using eigenvalues or eigenvectors:
$$
  A =
  \left[ {\begin{array}{ccc}
    1 & 1 & 1 \\
    0 & 2 & 2 \\
    0 & 0 & 3 \\
  \end{array} } \right]
$$
$$
  B =
  \left[ {\begin{array}{ccc}
    1 & 0 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 3 \\
  \end{array} } \right]
$$
I see that they are row-equivalent.  But how to determine the transition matrix $S$ such that $S^{-1}AS = B$?  I understand diagonalizability (which they are) but I am instructed not to use that concept.
I also know that I could set up a system of equations to solve for the entries of $S$ like $AS = SB$, but that strikes me as crude.
 [/INST] 
I also know that I could set up a system of equations to solve for the entries of $S$ like $AS=SB$, but that strikes me as crude.

Crude as it may be, it's certainly doable.
$$S = \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & j\end{bmatrix}$$
$$AS = \begin{bmatrix}a + d + g & b + e + h & c + f + j \\ 2 d + 2 g & 2 e + 2 h & 2 f + 2 j \\ 3 g & 3 h & 3 j\end{bmatrix}$$
$$SB = \begin{bmatrix}a & 2b & 3c \\ d & 2e & 3f \\ g & 2h & 3j\end{bmatrix}$$
From the bottom row, we have $3g = g$ and $3h = 2h$, which means $g = 0$ and $h = 0$.
$$\begin{bmatrix}a + d & b + e & c + f + j \\ 2 d & 2 e & 2 f + 2 j \\ 0 & 0 & 3 j\end{bmatrix} = \begin{bmatrix}a & 2b & 3c \\ d & 2e & 3f \\ 0 & 0 & 3j\end{bmatrix}$$
From the middle row, we have $2d = d \implies d = 0$, and $2f + 2j = 3f \implies f = 2j$.
$$\begin{bmatrix}a & b + e & c + 3j \\ 0 & 2 e & 6j \\ 0 & 0 & 3 j\end{bmatrix} = \begin{bmatrix}a & 2b & 3c \\ 0 & 2e & 6j \\ 0 & 0 & 3j\end{bmatrix}$$
And from the top row, we have $b + e = 2b \implies e = b$ and $c + 3j = 3c \implies 3j = 2c \implies j = \frac{2}{3}c$.
$$\begin{bmatrix}a & 2b & 3c \\ 0 & 2 e & 4c \\ 0 & 0 & 3 j\end{bmatrix} = \begin{bmatrix}a & 2b & 3c \\ 0 & 2e & 4c \\ 0 & 0 & 2c\end{bmatrix}$$
Putting this together, we have:
$$S = \begin{bmatrix}a & b & c \\ 0 & b & \frac{4}{3}c \\ 0 & 0 & \frac{2}{3}c\end{bmatrix}$$
Where $a$, $b$, and $c$ can be anything (except 0, if you want $S^{-1}$ to exist).  I'll pick $a = 1$, $b = 1$ and $c = 3$.
$$S = \begin{bmatrix}1 & 1 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 2\end{bmatrix}$$
</s>