Similar matrices have the same characteristic polynomial

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This proof is supposed to be easy but it may stun students who have not familiarized themselves with possible manipulations that may be used for such linear algebra questions.
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punditgi
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RyanLewis-Johnson-wqxs
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mpandejonasshane
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jamesharmon
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Characteristic Polynomial:determinANT(A-λ I) Char. Poly: det(A-λI)=0

RyanLewis-Johnson-wqxs
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Fact that similar matrices have the same eigenvalues is used in numerical linear algebra to find eigenvalues
There are used transformations which preserve similarity like Householder reflections or Givens rorations or just modification of Gaussian elimination
To preserve similarity we must perform elementary operations on both rows and columns

holyshit
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Given A=PBP^-1 W.T.S. det(A-λI)=det(B-λI) proof A-λI=PBP^-1-λI=PBP^-1= λPIP^-1(PIP^-1=I)

RyanLewis-Johnson-wqxs
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Similar matrices have the same characteristics polynomial A matrix A is similar to another matrix B if there exists matrix p such that A=PBP^-1 or B=P^-1AP

RyanLewis-Johnson-wqxs
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What does drawing the square in the end mean?

Anmol_Sinha
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PBP^-1-P λIP^-1=P(BP^-1-λIP^-1) A-λI=P(B-λI)P^-1

RyanLewis-Johnson-wqxs
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det(A-λI)=det(B-λI) det P^-1=1/(det P)

RyanLewis-Johnson-wqxs
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DiverseDose
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Why is lambda the same on both sides? The characteristic polonomial has some value lambda, but it doesnt need to be the same value for all matrices. Using the same lambda on both sides seems to assume the matrices have the same characteristic. I would think that needs to be proven.

inGuy
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It's easy to show that if A and B are similar then (A - λI) and (B - λI) are similar. Next use the fact that similarity preserves the determinant. That's it.

shacharh