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How are the matrices: $$A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&2\end{bmatrix},\qquad B=\begin{bmatrix}1&1&0\\0&1&0\\0&0&2\end{bmatrix}$$ fundamentally different? These are in Jordan form, so are not conjugate. They have the same eigenvalues, eigenvectors, and eigenspaces.

They have the same characteristic polynomial, and the minimal polynomials differ slightly $\mu_A=(t-1)(t-2)$ and $\mu_B=(t-1)^2 (t-2)$.

A matrix is classified uniquely, up to order of Jordan blocks, by the Jordan normal form. Over $\Bbb C$ is it true that the minimal polynomial classifies the matrix up to this ordering?

What other things are fundamentally different about these matrices?

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    $\begingroup$ You might find my answer here to be useful. In a nutshell: both $(A - I)$ and $(B - I)$ each have an invariant subspace (a "generalized eigenspace of 1") over which they are nilpotent. The difference is that one is zero over this subspace, whether the other acts as a non-trivial nilpotent operator. $\endgroup$ – Omnomnomnom Aug 16 '17 at 6:05
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    $\begingroup$ The minimal polynomial does not classify the matrix up to that ordering. Consider the matrices $\text{diag}(1,1,2)$ and $\text{diag}(1,2,2)$. They both have minimal polynomial $(t-1)(t-2)$. However, in two and three dimensions the minimal polynomial together with the characteristic polynomial is enough. In higher dimensions, though, even these together are not enough, and you need the "elementary divisors." $\endgroup$ – symplectomorphic Aug 16 '17 at 6:08
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    $\begingroup$ Also, your matrices $A$ and $B$ don't have the same eigenvectors. The vector $(0,1,0)^T$ is an eigenvector of $A$ with eigenvalue $1$, but it's not an eigenvector of $B$. $\endgroup$ – symplectomorphic Aug 16 '17 at 6:14
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The matrices \begin{align*} J_1 &=\left[\begin{array}{rr|r|r} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \hline 0 & 0 & 1 & 0 \\ \hline 0 & 0 & 0 & 1 \end{array}\right] & J_2 &=\left[\begin{array}{rr|rr} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \hline 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right] \end{align*} are in Jordan canonical form and satisfy $\chi_{J_1}(t)=\chi_{J_2}(t)=(t-1)^4$. Their minimal polynomials are $$ \mu_{J_1}(t)=\mu_{J_2}(t)=t^2-2\,t+1 $$ This shows that the minimal polynomial does not determine the Jordan canonical form of a matrix.

Now, let's consider your examples \begin{align*} A &= \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array}\right] & B &=\left[\begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array}\right] \end{align*} As you observe, these matrices have the same characteristic polynomials. Thus the eigenvalues of these matrices are $\lambda=1$ and $\lambda=2$. Moreover, the algebraic multiplicities of these eigenvalues are $\DeclareMathOperator{am}{am}\am_A(1)=\am_B(1)=2$ and $\am_A(2)=\am_B(2)=1$.

However, the geometric multiplicities are $\DeclareMathOperator{gm}{gm}\gm_A(1)=2\neq 1=\gm_B(1)$ and $\gm_A(2)=\gm_B(2)=1$.

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