A few proofs in ranks and null

**tttcomrader** · April 25th 2008, 06:11 AM

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

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

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

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

**ThePerfectHacker** · April 25th 2008, 08:37 AM

Originally Posted by tttcomrader

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

I start you off. Let $x\in N(U)$ then by definition (I assume it is the nullspace) $U(x) = \bold{0}\implies U(U(x)) = U(\bold{0}) = \bold{0}$ . Thus, $x\in N(U^2)$ . Now generalize the argument to prove that $N(U^{k}) \subseteq N(U^{k+1})$ .

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

Using the rank nullity theorem we find that $\text{rank}(U^m) + \text{nullity}(U^m) = \text{rank}(U^{m+1})+\text{nullity}(U^{m+1})$ thus, $\text{nullity}(U^m) = \text{nullity}(U^{m+1})$ . Thus, $\text{dim}N(U^m) = \text{dim}N(U^{m+1})$ since $N(U^m)$ is a subspace of $N(U^{m+1})$ it actually means $N(U^m) = N(U^{m+1})$ (because their dimensions are the same). Now generalize this argument for $m+2,m+3,...$ .
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Sorry I cannot help with (c), my linear algebra is not that great.

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