Show that matrices are not similar I have to show that the following matrices are not similar:
$$A = \left[\begin{matrix} 1 & 3 & -3 \\ -3 & 7 & -3 \\ -6 & 6 & -2\end{matrix}\right]$$
and $$A' = \left[\begin{matrix} 5 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 3\end{matrix}\right]$$
I know that 2 matrices $A$ and $A'$ are similar if there exists an invertible matrix $B$, such that $$A'= B^{-1}AB$$
According to Wikipedia, similar matrices share some properties (for example they have the same eigen values), but I don't know how to start, since I have missed the last lectures of my linear algebra course, unfortunetely. Unfortuntely, life has not been so completely fair with me :(
Should I just check if they have the same eigen values?
 A: $\det A = -32, \det A' = 60$. Similar matrices have the same determinant.
A: Does the above method always work? For example, if you have 2 matrices in Jordan form, the determinants can both be 0 (say, all 0 along their diagonals). In this case, the 2 matrices can be NOT similar but still share the same determinant, making the above proof not necessarily true for all cases?
A: Since adding numbers along the diagonal is simple, you can start with the fact that the trace is preserved by similarity transformations, i.e. $\text{tr}(A) = \text{tr}(B^{-1}AB)$. 
Do $A$ and $A'$ have the same trace? If not, then $A$ and $A'$ are not similar.
EDIT: The trace of a square matrix is simply the sum of the entries on the diagonal, i.e. for an $n \times n$ matrix $A$, we have $\text{tr}(A) = \displaystyle\sum_{i = 1}^{n}A_{i,i}$. 
A: After some deeper understanding of these concepts, I came up with my answer to this question.
We know that $A$ and $A'$ are similar if there exists an invertible matrix $B$ (there must exist $B^{-1}$), such that $$A' = B^{-1}AB$$
Since $A'$ must be equal to $B^{-1}AB$, then also $$det(A') = det(B^{-1}AB) = det(B^{-1})\cdot det(A)\cdot det(B)$$
Since $B^{-1}$ is the inverse of $B$, then $$det(B^{-1}) = \frac{1}{det(B)}$$
We can then replace this in the previous expression 
$$det(A') = det(B^{-1}AB)= det(B^{-1})\cdot det(A)\cdot det(B) = \frac{1}{det(B)}\cdot det(A)\cdot det(B) = det(A)$$
Which basically means that if $A$ and $A'$ are similar, then $$det(A) = det(A')$$
If we now calculate the determinant of $A$ and the determinant $A'$, we have different results, so $A$ is not similar to $A'$.
