I want to show that these two matrices are similar
$$\begin{pmatrix}
6 & 6 & -15 \\
1 & 5 & -5 \\
1 & 2 & -2
\end{pmatrix} \ \ \ \ \ \ \begin{pmatrix}
37 & -20 & -4 \\
34 & -17 & -4 \\
119 & -70 & -11
\end{pmatrix}$$
It is from my homework. I have showed that their characteristic polynomial are the same, $f(x)=(x-3)^3$, but I know that it isn't sufficient. How do I proceed?
In class, we not studying yet the Jordan canonical form.
Thank you very much!
Best Answer
Without really explaining Jordan canonical forms, consider:
To be similar, $P$ must exist such that: $$A = P^{-1}BP$$
Let $u_i$ be the eigenvectors of $A$ and $v_i$ be the eigenvectors of $B$. Define two matrices $G$ and $H$ with those eigenvectors as columns:
$$ G = \left( \begin{array}{ccc} u_1 & u_2 & ... \end{array} \right),\ \ \ H = \left( \begin{array}{ccc} v_1 & v_2 & ... \end{array} \right) $$
Since the eigenvectors are linearly independent, we know both $G^{-1}$ and $H^{-1}$ exist. Use these facts to come up with two new matrices: $$ S_A = G^{-1}AG, \ \ \ S_B = H^{-1}BH $$
$S_A$ is similar to $A$ and $S_B$ is similar to $B$.
If $S_A = S_B$, then: $$ S_A = S_B\\ G^{-1}AG = H^{-1}BH\\ AG = GH^{-1}BH\\ A = GH^{-1}BHG^{-1} $$ With $P=HG^{-1}$ and $P^{-1}= GH^{-1}$, $A$ and $B$ are similar: $$A = P^{-1}BP$$ Therefore, you just need to construct $G$ and $H$ from the eigenvectors to find $P$ and show that they're similar.