# Rank and number of distinct eigenvalues

I have this question:

Let $$A\in M_{10}$$ be a matrix with $$5$$ distinct eigenvalues. Is $$\operatorname{rank}(A) \geq 5$$?

My approach is the following:

Since matrix $$A$$ has $$5$$ distinct eigenvalues, only one of those five can be equal to zero. In a worst-case scenario, all of the other five eigenvalues are also equal to zero. That means that the matrix has 6 eigenvalues equal to zero, which is equivalent to $$\dim(\ker(A))=6$$. Now, using the rank-nullity theorem, we conclude that $$\operatorname{rank}(A)=4$$, which means that the statement is not true.

Is my approach correct? If it isn't, what would be the answer?

Yes, that's correct. And if you want a specific counterexample then you can take a diagonal matrix with $$4$$ different non zero entries and $$6$$ zeros on the main diagonal.