# Thread: A few proofs in ranks and null

1. ## A few proofs in ranks and null

Let U be a linear operator on a finite dimensional vector space V, prove that:

a) $\displaystyle N(U) \subseteq N(U^2) \subseteq N(U^3) \subseteq ... \subseteq N(U^k) \subseteq N(U^k+1) \subseteq . . .$

b) If $\displaystyle rank (U^m) = rank (U^{m+1})$ , then $\displaystyle N(U^m) = N(U^k) \ \ \ \forall k \geq m$

c) Let T be linear operator on V whose characteristic polynomial splits, and let $\displaystyle \lambda _1 \lambda _2 , . . . , \lambda _k$ be distinct eigenvalues of T. Then T is diagonalizable if and only if $\displaystyle rank (T- \lambda _i I ) = rank ((T- \lambda _i I )^2 )$ for $\displaystyle 1 \leq i \leq k$

a) $\displaystyle N(U) \subseteq N(U^2) \subseteq N(U^3) \subseteq ... \subseteq N(U^k) \subseteq N(U^k+1) \subseteq . . .$
I start you off. Let $\displaystyle x\in N(U)$ then by definition (I assume it is the nullspace) $\displaystyle U(x) = \bold{0}\implies U(U(x)) = U(\bold{0}) = \bold{0}$. Thus, $\displaystyle x\in N(U^2)$. Now generalize the argument to prove that $\displaystyle N(U^{k}) \subseteq N(U^{k+1})$.
b) If $\displaystyle rank (U^m) = rank (U^{m+1})$ , then $\displaystyle N(U^m) = N(U^k) \ \ \ \forall k \geq m$
Using the rank nullity theorem we find that $\displaystyle \text{rank}(U^m) + \text{nullity}(U^m) = \text{rank}(U^{m+1})+\text{nullity}(U^{m+1})$ thus, $\displaystyle \text{nullity}(U^m) = \text{nullity}(U^{m+1})$. Thus, $\displaystyle \text{dim}N(U^m) = \text{dim}N(U^{m+1})$ since $\displaystyle N(U^m)$ is a subspace of $\displaystyle N(U^{m+1})$ it actually means $\displaystyle N(U^m) = N(U^{m+1})$ (because their dimensions are the same). Now generalize this argument for $\displaystyle m+2,m+3,...$.