The geometric multiplicity is equal to the algebraic multiplicity, then t is diagonalizable.
The word "nilpotent" is missing from the title, rather confusing, since being nilpotent is easily seen to be the only possible obstruction against being diagonalisable for a rank 1 operator. After all, by rank-nullity this implies the nullity equals n−1, and this is (by definition) the dimension of the eigenspace for 0. Also Im(T) is always T-stable, so being of dimension 1 its nonzero vectors are eigenvectors, and if this is for a nonzero eigenvalue λ, then the eigenspaces for 0 and for λ are complementary, and so T is diagonalisable.
So the only thing that can go wrong is that the nonzero vectors of Im(T) are eigenvectors for 0, but then clearly T2=0 so T is nilpotent.
It would have been a bit more fun if instead of T not being nilpotent it had been given that T has nonzero trace. Since 0 is a root of the characteristic polynomial χT with multiplicity at least n−1 (which is its geometric multiplicity), the "final" root of χT equals the sum of all its roots, which is the trace of T. This is also the eigenvalue of nonzero vectors in Im(T), and as we have seen it is zero if and only if T is nilpotent if and only if T is not diagonalisable.
Hence the answer is the geometric multiplicity is equal to the algebraic multiplicity, then t is diagonalizable.
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