I have seen the following True/ False question
If $A ∈ M_n(\mathbb{R})$ (with $n \geq 2$) has rank 1, then the minimal polynomial of $A$ has degree 2.
The statement is true. If we generalize the conditions a little bit, then the following question occur:
If $A ∈ M_n(\mathbb{R})$ (with $n \geq 2$) has rank $k < n$, then what will be the degree of the minimal polynomial of A?
Do the degree must be $k+1$?
A linear operator$~\phi$ on an $n$-dimensional space $V$ having rank $k$ just means that its image subspace $W=\phi(V)$ has $\dim{W}=k$. Since the image is always $\phi$-stable we can restrict $\phi$ to $W$, but one has essentially no information about this restriction, except that the dimension of its kernel cannot exceed the codimension $n-k$ of $W$ in $V$ (by rank-nullity, since the rank of the unrestricted $\phi$ cannot exceed the rank of $\phi|_W$ by more than that). So for any polynomial$~P$, one has $(PX)[\phi]=0$ if and only if $P[\phi|_W]=0$, and assuming $k<n$ (so that the minimal polynomial $\mu\in K[X]$ of$~\phi$ has at least one factor$~X$), the minimal polynomial $\mu'$ of $\phi|_W$ satisfies $\mu=\mu'X$ and therefore $\deg\mu=\deg\mu'+1$.
Now for $1=k<n$ one necessarily has $\deg\mu'=1$ (the degree of a minimal polynomial of an an operator on a finite non-zero dimensional space lies between $1$ and the dimension, inclusive) and hence $\deg\mu=2$.
But for $1<k<n$ all one can say is that $1\leq\deg\mu'\leq k$ (for the same reason; the mentioned limit on the kernel dimension does not prevent us from attaining any of these values, as that can be done with (operators defined using) just triangular matrices with entries $1$ on the diagonal), so all one can say is $2\leq\deg\mu\leq k+1$.