I think of a counterexample to show that $3 \times 3$-matrices $A$ and $B$ can have the same characteristic polynomials and minimal polynomials without being similar:

characteristic polynomials are both $(t-2)^3$.

one minimal polynomial is $(t-2)^2 (t-2)$ and another is $(t-2)(t-2)(t-2)$.

Then, I get the following Jordan forms of linear transformations for $A$ and $B$ respectively:

\begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0&0&2 \end{bmatrix}


\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0&0&2 \end{bmatrix}

Therefore, $A$ and $B$ are not similar. Is it correct?

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    $\begingroup$ You might want to review the definition of minimal polynomial. The minimal polynomial of your second matrix is $t-2$, not $(t-2)^3$. $\endgroup$ – amd Feb 2 '18 at 3:22
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    $\begingroup$ Just an additional comment, in my memory if there are two $3\times 3$ matrices with same characteristic polynomial and minimal polynomial, then they must be similar (something using degree argument). Please correct me if I'm wrong. $\endgroup$ – ElfHog Feb 2 '18 at 3:29

Your example is incorrect: the minimal polynomial of $B$ is simply $t - 2$.

When constructing such a counterexample, it is helpful to note that the exponent is the minimal polynomial is the size of the largest associated Jordan block, while the exponent in the characteristic polynomial is the overall multiplicity of the eigenvalue. So, for example, the matrices $$ \pmatrix{2&1\\&2\\&&2&1\\&&&2\\}, \quad \pmatrix{2&1\\&2\\&&2\\&&&2} $$ Both have characteristic polynomial $(t-2)^4$ (since $2$ is their only eigenvalue) and minimal polynomial $(t - 2)^2$ (since their largest Jordan block is of size $2$).

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    $\begingroup$ also, jordan forms for matrices of dimension 3 are completely determined by the minimal polynomial $\endgroup$ – qbert Feb 2 '18 at 3:29

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