# 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?

• Can you find the eigenvalue of $A'$? (Yes, you can check that). – user99914 May 19 '15 at 2:41
• Well, I can assume the determinant of $\lambda I - A' = 0$ and then solve for $\lambda$ (eigen values)? Something tells me that $A'$ is special because it's a diagonal matrix... – nbro May 19 '15 at 2:42

## 3 Answers

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 $\text{tr}(A) = \displaystyle\sum_{i = 1}^{n}A_{i,i}$.

$\det A = -32, \det A' = 60$. Similar matrices have the same determinant.

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 $A'$.