Is upper Hessenberg form preserved through similarity transformation Suppose $X$ is non-singular and $M$ is upper Hessenberg. 
Is $X^{-1}MX$ also upper Hessenberg. 

No.
Let
$$M
= \begin{pmatrix}
1 & 4 & 2 & 3 \\
3 & 4 & 1 & 7 \\
0 & 2 & 3 & 4 \\
0 & 0 & 1 & 3 \\
\end{pmatrix}
~~~~~~
X = 
\begin{pmatrix}
 1 & 0 & 0 & 1 \\
 1 & 2 & 3 & -1 \\
 1 & 2 & -1 & 0 \\
 1 & 1 & 1 & 1
\end{pmatrix}
$$
then
$$X^{-1} =
\begin{pmatrix}
 \frac{11}{7} & \frac{3}{7} & \frac{1}{7} & -\frac{8}{7} \\
 -\frac{6}{7} & -\frac{1}{7} & \frac{2}{7} & \frac{5}{7} \\
 -\frac{1}{7} & \frac{1}{7} & -\frac{2}{7} & \frac{2}{7} \\
 -\frac{4}{7} & -\frac{3}{7} & -\frac{1}{7} & \frac{8}{7}
\end{pmatrix}
$$
and we have
$$X^{-1}MX =
\begin{pmatrix}
 \frac{132}{7} & \frac{190}{7} & \frac{188}{7} & -\frac{4}{7} \\
 -\frac{37}{7} & -\frac{54}{7} & -\frac{72}{7} & \frac{13}{7} \\
 -\frac{5}{7} & -\frac{16}{7} & -\frac{5}{7} & \frac{8}{7} \\
 -\frac{62}{7} & -\frac{85}{7} & -\frac{97}{7} & \frac{4}{7}
\end{pmatrix}$$
which is not upper Hessenberg.
