Let $A$ be an $n \times n $ matrix such that the set of all nonzero eigenvalues has exactly $r$ elements. Which of the following statements is true?
(a) rank$(A)\le r$
(b) If $r=0$, then rank$(A) <n-1$
(c) rank$(A) \ge r$
(d) $A^2$ has $r$ distinct nonzero eigenvalues
$\bf{Try:}$
Let choose $A=\begin{pmatrix} 1& 0 & 0\\0 & - 1 & 0\\ 0 & 0 & 1\end{pmatrix}$
Hence the set of all nonzero eigenvalues of $A$ is $\{1,-1\}$.Hence $r=2$ but rank$(A) =3$. Hence option (a) is false.
Now $A^2=I_{3\times 3} $ . But $A^2$ has only one nonzero eigenvalue hence $r=1$. Hence option (d) is also false.
Now consider the matrix $A=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix} $
For this matrix $r=0$ but it has rank $1$.Hence option (b) also fslse. So option (c) must be true. But can this option be proved without checking by example?