A is Mn×n(C) with rank r and m(t) is the minimal polynomial of A. Prove deg $m(t) \leq r+1$ $A$ is a matrix of $M_{n \times n}(\mathbb{C})$ with rank $r$ and $m(t)$ is the minimal polynomial of A.
I need to prove that :


*

*deg $m(t) \leq r+1$

*I need to find a condition of the matrix A, in which deg $m(t) = r+1$


Can anyone help me ? The solution involves the primary decomposition theorem for matrices and Jordan form..
 A: No primary decompositions or Jordan forms are needed, the fact that this is over the complex numbers is irrelevant. One may assume $r<n$ (without which the question is pointless, and there would be no way to have 2.). Then $0$ is an eigenvalue, and $t$ is a factor of $m\in\Bbb C[t]$.
Now for any factor $f$ of $m$, one has that the quotient $q=m/f$ equals the minimal polynomial$~m'$ of the restriction$~\phi|_W$ of the linear operator$~\phi$ (represented by the given matrix$~A$) to the image$~W$ of $f[\phi]$. This is because on one hand $0=m[\phi]=q[\phi]\circ f[\phi]$, so $q[\phi|_W]=0$ (on $W$), and on the other hand $m'$ cannot be a strict divisor of$~q$, or else $(fm')[\phi]=0$ would contradict the minimality of$~m$.
Apply this with $f$ the factor $t$ of $m$, and it says that $m'=m/t$ is the minimal polynomial of the restriction of$~\phi$ to the image$~W=\phi(V)$. By definition of the rank $\dim(W)=r$, and so $\deg(m')\leq r$ (by the Cayley-Hamilton theorem, if you like), which gives 1., namely $\deg(m)\leq r+1$.
For point 2., the natural (necessary and sufficient) condition is that the restriction of$~\phi$ to $\phi(V)$ has minimal polynomial of maximal degree, namely$~r$. I'm not sure that is the condition they are after. Various sufficient conditions can be given instead, for instance that $A$ has $r$ nonzero eigenvalues. Or weaker, that all eigenspaces for nonzero eigenvalues have dimension$~1$, while the eigenspace for $0$ (which may of course have larger dimension) intersects$~W=\phi(V)$ with dimension at most$~1$. Stated differently, at most one Jordan block for $\lambda=0$ has size larger than$~1$. (This second sufficient condition does depend on working over$~\Bbb C$, or at least on the condition that $m$ splits into linear factors.)
