TITLE: what can be derived from similar matrix
QUESTION [5 upvotes]: If $A=\begin{pmatrix}
0&\star&\star \\
\star&x&\star \\
\star & \star & 5
\end{pmatrix}$ is similar to $B=\begin{pmatrix}
1&0&0 \\
0&y&0 \\
0 & 0 & 10
\end{pmatrix}$,where $\star$ represents unknown number. 
Find $x$ and $y$

I know, the trace of $A$ is equal to the trace of $B$, in other words, $5+x=11+y$, and their determinant and characteristic polynomial is also the same. but $\det A$ and $\det(\lambda I -A)$ have a lot of unknown quantity. To what degree we can determine $A$ from the condition?

REPLY [2 votes]: There is no further relation between $x$ and $y$ appart from $5+x=11+y$ (equality of traces). The key point is to have in mind that the matrices will be similar if the trace, the determinant and the $\lambda$ factor in the characteristic polynomial are the same (the sum of the three $2\times 2$ subdeterminants); it is not a complete equivalence (think for instance in non-diagonalisable matrices), but this case will only occur for a finite number of combinations of the stars.
So the strategy is as follows: choose $y\in \mathbb{R}$ and set $x=6+y$. Then the stars may be given values so that 

The determinant of $A$ is $10y$
The sum of the three $2\times 2$ subdeterminants (with appropriate sign) is $y+10y+10$.

Then $y$ may be any real number.