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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?

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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.

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