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

Question

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?

Answer 1

• $A$ 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.

Answer 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 $$(CD)^{-1}=D^{-1}C^{-1}$$

Answer 3

So you are given that the matrix is similar to a matrix which is similar to a diagonal matrix (diagonalizable). But similarity is a transitive property, as is easily checked: $A=PBP^{-1}$ and $B=QCQ^{-1} \implies A=(PQ)C(PQ)^{-1}$. The result follows.