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Are the matrices $A = \begin{bmatrix} 1 & 3 & 3\\0 & 4 & 5\\0 & 0 & 6\end{bmatrix}$ and $B = \begin{bmatrix} 6 & 7 & 8\\0 & 4 & 9\\0 & 0 & 1\end{bmatrix}$ similar?

I just learned how to show that matrices are similar from a friend and want to make sure I have the right idea for this problem.

My first step is always to check that the eigenvalues are the same since similar matrices have the same eigenvalues. That doesn't help here.

Since the eigenvalues for $A$ are distinct, it is a simple matrix and thus semi-simple as well. I know a matrix is semi-simple if and only if it is diagonalizable so $A = P^{-1}DP$ for some matrix $P$ where $D$ is the diagonal matrix corresponding to $A$. The same applies for $B$ so $B = Q^{-1}DQ$ for some matrix $Q$ where $D$ is the same diagonal matrix as before since $A$ and $B$ have the same eigenvalues.

This implies $PAP^{-1} = D$ so $B = Q^{-1}(PAP^{-1 })Q = (Q^{-1}P)A(Q^{-1}P)^{-1}.$ Thus $A$ and $B$ are similar.

Is this the best way to approach this problem?

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It is the best way, but in my opinion you could have put it in a simpler way. Whenever you have a $n\times n$ upper triangular matrix with $n$ distinct entries $\lambda_1,\lambda_2,\ldots,\lambda_n$ in the main diagonal, $A$ is similar to the diagonal matrix such that the entries of the main diagonal are $\lambda_1,\lambda_2,\ldots,\lambda_n$. In your case, this shows that $A$ and $B$ are similar to$$\begin{bmatrix}1&0&0\\0&4&0\\0&0&6\end{bmatrix}\text{ and to }\begin{bmatrix}6&0&0\\0&4&0\\0&0&1\end{bmatrix}$$respectively. Since these two matrices are clearly similar, $A$ and $B$ are similar too.

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