Let T: VV be a linear operator on a finite-dimensional F-vector space V, and let f(t) be the characteristic polynomial of T. Show that T is diagonalizable if and only if f(t) splits over F and rank(T - XI) = rank((T – XI)²) for every eigenvalue A of T. (You can use the theorem that T has a Jordan basis iff f(t) splits, and the theorem that says nullity ((T – XI)²) – nullity (T – XI) = ···)

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