How do I prove that the 2 ranks are equivalent

Asked Oct 31 '19 at 15:15

Active Oct 31 '19 at 15:29

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Let A be a diagonalizable matrix and $\lambda$ an eigenvalue of A $\DeclareMathOperator{\rank}{rank}$ Show that $$\rank(\lambda I - A)= \rank((\lambda I - A )^2)$$

Iam not sure whether this is allowed but i tried premultiplying by P and P^{-1} such that

$$\rank(P(\lambda I - A)P^{-1})= \rank(\lambda PIP^{-1} -PAP^{-1}) $$

but is this even correct?

edited Oct 31 '19 at 15:29

Bernard

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asked Oct 31 '19 at 15:15

user718896

I meant P^-1 as inverse of P i dont know why it didnt format correclty – user718896 Oct 31 '19 at 15:18
1

You have to write P^{-1}. – parsiad Oct 31 '19 at 15:19
Do you know the rank-inequalities of Sylvester? They state that $\text{rank}(AB) \le \min(\text{rank}(A), \text{rank}(B))$, which gives one of the two desired inequalities. – ViktorStein Oct 31 '19 at 15:20
@ViktorGlombik no sorry i havent learn that – user718896 Oct 31 '19 at 15:22
It's easy to prove, though, see i.e this answer on Math.SE. – ViktorStein Oct 31 '19 at 15:25
@ViktorGlombik oh so would something like rank((\lambda I - A)(\lambda I - A) <= min(rank((\lambda I - A),rank((\lambda I - A)) – user718896 Oct 31 '19 at 15:33
Yes, that is correct. – ViktorStein Oct 31 '19 at 15:40
Oh ok sorry when i googled search my question that result didnt appear – user718896 Oct 31 '19 at 15:57

How do I prove that the 2 ranks are equivalent

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