If $A$ is an $n$-by-$n$ matrix with complex entries, (i.e., $A\in M\_n(\mathbb{C})$,) $A$ must have $n$ eigenvalues, counting algebraic multiples. But it is not always true that $A$ has $n$ linearly independent eigenvectors. So, what necessary and sufficient condition may be add, to ensure that $A$ has $n$ linearly independent eigenvectors?

Of course, the simpler the better.

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I have another related question:

I can't figure out why, intuitively, that algebraic multiple doesn't mean more than one linearly independent eigenvectors. I mean in my tuition, a multiple appear **because** there is a subspace with dimension>1 being scaled "evenly" in every direction. If the multiple is 2, how come I may fail to find 2 linearly independent vectors in this subspace?