If A is diagonalizable and B is similar to A, then B is diagonalizable

by Soon_to_be_code_master   Last Updated April 16, 2018 07:20 AM

here is my proof, i need someone to check it for me.

$A = PDP^{-1}$
$B = PAP^{-1}$

Now i plug in A in B

$B = P(PDP^{-1})P^{-1}$

now i cancel out one of the $P$ and $P^{-1}$

$B = PDP^{-1}$

is this correct?



Answers 2


The $P$ in both equation need not be the same

Let $$A = PDP^{-1}$$ $$B = QAQ^{-1}$$

and repeat your argument.

Useful property:

If $C$ and $D$ are invertible, then $$(CD)^{-1}=D^{-1}C^{-1}$$

Siong Thye Goh
Siong Thye Goh
April 16, 2018 07:02 AM

  • $A$ is diagonizable: $A=PDP^{-1}$.

  • $B$ is similar to $A$: $B=QAQ^{-1}$.

Hence

$$B=Q(PDP^{-1})Q^{-1}=(QP)D(P^{-1}Q^{-1})=(QP)D(QP)^{-1}$$ which establishes the claim.

Yves Daoust
Yves Daoust
April 16, 2018 07:14 AM

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