Similarity of $3\times3$ matrices via Jordan canonical form Let matrices
$$A=\begin{bmatrix}
-3&amp;3&amp;-2\\
-7&amp;6&amp;-3\\
1&amp;-1&amp;2
\end{bmatrix},\quad 
B=\begin{bmatrix}
0&amp;1&amp;-1\\
-4&amp;4&amp;-2\\
-2&amp;1&amp;1
\end{bmatrix},\quad
C=\begin{bmatrix}
0&amp;-1&amp;-1\\
-3&amp;-1&amp;-2\\
7&amp;5&amp;6
\end{bmatrix},\quad
D=\begin{bmatrix}
0&amp;1&amp;2\\
0&amp;1&amp;1\\
0&amp;0&amp;2
\end{bmatrix}
$$
My goal is to determine which of them are similar to each other. First, we observe that
$$\det(A)=\det(B)=\det(C)=4\neq\det(D)=0$$
Thus, $D$ cannot be similar to any of $A,B,C$. $\det(\lambda I-A)=0$ and similarly for $B,C$. Denote $\sigma(X)=\operatorname{spectrum of matrix }X$. With a slight abuse of notation, to denote that $\lambda_2=2$ has algebraic multiplicity of $2$, I found
$$\sigma(A)=\sigma(B)=\sigma(C)=\{\lambda_1=1,\lambda_2=2,\lambda_2=2\}$$
This means that all of them have the same Jordan Canonical Form which is (up to permutations of the diagonal Jordan blocks)
$$
J_A=J_B=J_C=\begin{bmatrix}
\lambda_1&amp;0&amp;0\\
0&amp;\lambda_2&amp;1\\
0&amp;0&amp;\lambda_2
\end{bmatrix}
=
\begin{bmatrix}
1&amp;0&amp;0\\
0&amp;2&amp;1\\
0&amp;0&amp;2
\end{bmatrix}
$$
Hence, $A,B,C$ are similar.
Is this justification correct? If yes, is there any faster way/trick to get to this result? Finding the eigenvalues by solving a 3-rd degree equation took me some time.
$B$ diagonalizes

$$
\frac{1}{2}
\left(
\begin{array}{rrr}
4 &amp; -2 &amp; 2 \\
-4 &amp; 3 &amp; -2 \\
-2 &amp; 1 &amp; 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 &amp; 1 &amp; -1 \\
2 &amp; 2 &amp; 0 \\
1 &amp; 0 &amp; 2 \\
\end{array}
\right) = I
$$
$$
\frac{1}{2}
\left(
\begin{array}{rrr}
4 &amp; -2 &amp; 2 \\
-4 &amp; 3 &amp; -2 \\
-2 &amp; 1 &amp; 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
0 &amp; 1 &amp; -1 \\
-4 &amp; 4 &amp; -2 \\
-2 &amp; 1 &amp; 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 &amp; 1 &amp; -1 \\
2 &amp; 2 &amp; 0 \\
1 &amp; 0 &amp; 2 \\
\end{array}
\right) = 
\left(
\begin{array}{rrr}
1 &amp; 0 &amp; 0 \\
0 &amp; 2 &amp; 0 \\
0 &amp; 0 &amp; 2 \\
\end{array}
\right)
$$

Regardless, this was going to require a fair amount of work by hand. I am not so much a fan of doing things quickly as of doing things well and including alternate methods that may indicate errors. In this case, one may confirm that $B^2 - 3 B + 2 I = 0.$ For $A$ and $C$ the minimal polynomial agrees with the characteristic polynomial, which means that each eigenvalue occurs in a single Jordan block, and since $2$ has higher multiplicity this means they do not diagonalize. In contrast, for $B$ we see that $2$ must occur in more than one Jordan block. In fact, there are two, each dimension one, meaning there is a basis of two eigenvectors for eigenvalue $2.$ I have typed in one way  to diagonalize $B.$ The matrix on the right has columns that are all eigenvectors
$B$ diagonalizes

$$
\frac{1}{2}
\left(
\begin{array}{rrr}
4 &amp; -2 &amp; 2 \\
-4 &amp; 3 &amp; -2 \\
-2 &amp; 1 &amp; 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 &amp; 1 &amp; -1 \\
2 &amp; 2 &amp; 0 \\
1 &amp; 0 &amp; 2 \\
\end{array}
\right) = I
$$
$$
\frac{1}{2}
\left(
\begin{array}{rrr}
4 &amp; -2 &amp; 2 \\
-4 &amp; 3 &amp; -2 \\
-2 &amp; 1 &amp; 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
0 &amp; 1 &amp; -1 \\
-4 &amp; 4 &amp; -2 \\
-2 &amp; 1 &amp; 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 &amp; 1 &amp; -1 \\
2 &amp; 2 &amp; 0 \\
1 &amp; 0 &amp; 2 \\
\end{array}
\right) = 
\left(
\begin{array}{rrr}
1 &amp; 0 &amp; 0 \\
0 &amp; 2 &amp; 0 \\
0 &amp; 0 &amp; 2 \\
\end{array}
\right)
$$

Regardless, this was going to require a fair amount of work by hand. I am not so much a fan of doing things quickly as of doing things well and including alternate methods that may indicate errors. In this case, one may confirm that $B^2 - 3 B + 2 I = 0.$ For $A$ and $C$ the minimal polynomial agrees with the characteristic polynomial, which means that each eigenvalue occurs in a single Jordan block, and since $2$ has higher multiplicity this means they do not diagonalize. In contrast, for $B$ we see that $2$ must occur in more than one Jordan block. In fact, there are two, each dimension one, meaning there is a basis of two eigenvectors for eigenvalue $2.$ I have typed in one way  to diagonalize $B.$ The matrix on the right has columns that are all eigenvectors