= Solution
The <minimal polynomial> $m_A$ is the <monic polynomial> of smallest degree satisfying $m_A(A)=0$. An annihilating <polynomial> exists, for example because $I,A,\ldots,A^{n^2}$ are linearly dependent in the $n^2$-dimensional space of <matrices>. A nonzero constant cannot annihilate $A$, so the minimal degree is positive.
If $p$ and $q$ are two such <monic polynomials> of minimal degree $d$, then $(p-q)(A)=0$ and $\deg(p-q)<d$ unless $p=q$. Minimality therefore forces $p=q$. <Polynomial> division also shows that $m_A$ divides every annihilating <polynomial>: the remainder after division would otherwise be an annihilator of smaller degree. This proves uniqueness rather than merely choosing one of several minimal-degree <polynomials>.
If $A$ is real, coefficientwise <complex conjugation> of $m_A(A)=0$ gives $\overline{m_A}(A)=0$. The conjugated <polynomial> is monic of the same degree, hence uniqueness gives $\overline{m_A}=m_A$. \b[The <minimal polynomial> of a real <matrix> has real coefficients.]
For $n\ge3$, use the <block diagonal matrix>
$$
\boxed{A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\oplus(-1)\oplus I_{n-3}.}
$$
The two-dimensional <Jordan block> is annihilated by $(t-1)^2$ but not by $t-1$, while the scalar block requires the factor $t+1$. A <polynomial> annihilates the <direct sum> exactly when it annihilates every block. Thus its <minimal polynomial> is the <least common multiple> $(t-1)^2(t+1)$. For $n=3$ the final identity block is omitted.
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