Prove/Disprove: Rank(B)=Rank(A) if B is Similar to A

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  • #1
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Homework Statement


Prove or give a counterexample:

If b is similar to A, then rank(B)=rank(A).


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The Attempt at a Solution


This is obvious when A and B have maximum rank (take the determinant of both sides of the similarity relation). My intuition and all the examples I have looked at tell me it is also true when they have less than maximum rank. But how to prove it? Is it easier to look at the nullity and use the Nullity-Rank Theorem?
 
  • #2
Since B is similar to A, then by definition of singularity, there is an invertible nxn matrix such that PB=AP or [tex]B=P^{-1}AP[/tex]

Let rank(B)=c, then what can you say about rank(PB) and rank([tex]P^{-1}AP[/tex] ) compared with c? Then, you can use the fact that similarity is an equivalence relation to show that rank(B)=rank(A)
 
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  • #3
Since B is similar to A, then by definition of singularity, there is an invertible nxn matrix such that PB=AP or [tex]B=P^{-1}AP[/tex]

Let rank(B)=c, then what can you say about rank(PB) and rank([tex]P^{-1}AP[/tex] ) compared with c?

Yes, I was trying to go that route. I was trying to remember whether an invertible matrix always preserves rank. I couldn't find that theorem in my book...is it true?
 

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