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Is the chain of implications below correct or does it break down somewhere along the way?

Let $A$ be a square matrix, $D$ a diagonal matrix and $P$ an eigenvector matrix. Consider $A = P^{-1}DP$ which implies $A = PDP^{-1}$ and so $A = P^{-1}DP = PDP^{-1}.$ Then $A = Q^{-1}DQ = QDQ^{-1}$ which implies $A = Q^TDQ = QDQ^T$ since $Q^T = Q^{-1}.$

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    The first implication doesn't hold in general. And how does $Q$ relate to $P$ or $A$? – 2017-02-11
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    @ Git Gud, For the first implication I used the following proposition: https://s23.postimg.org/mowa6tvdn/333.png Maybe I misused it? – 2017-02-11
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    You did. One of the assumptions in the proposition is $P^{-1}AP=D$, or equivalently $A=PDP^{-1}$. In this problem you start from $A = P^{-1}DP$. See the difference? – 2017-02-12
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    @ Git Gud, I guess the problem I'm having here is the definition of similarity of matrices. If I say we can swap $A, D$ in $P^{-1}AP = D$ and have the equality still hold, in what ways am I violating the definition of similarity of matrices? – 2017-02-12
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    You wouldn't be violating the definition of similarity in anyway. The thing is, you being able to swap $A$ and $D$ and have the equality still hold just doesn't happen (not often). – 2017-02-12
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    One last question, $P^{-1}AP = D$ and $A = P^{-1}DP$ say the same thing, namely, $A, B$ are similar, right? How come these equalities are not equivalent, then? – 2017-02-12
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    I think it's better to say things with a little more detail. Given two square matrices, $A$ and $D$, one says that $A$ is similar to $D$ if, and only if, $\exists P(P^{-1}AP=D)$. Symmetrically, one says that $D$ is similar to $A$ if, and only if, $\exists Q(Q^{-1}DQ=A)$. It can be proved that if $A$ is similar to $D$, then $D$ is similar to $A$ and so one adopts the terminology "$A$ and $D$ are similar". You're basically operating under the assumption that $P=Q$. This isn't necessarily true. Try it with a [specific example](http://www.wolframalpha.com/input/?i=%7B%7B1,2%7D,%7B0,3%7D%7D). – 2017-02-12

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One way to see that your reasoning is wrong is that, if it were true, then every diagonalizable matrix would be symmetric (because if $D$ is diagonal then $D=D^T$), which is clearly wrong.

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    @ Fimpellizieri, The problem I am having is if $P^{-1}AP = D$ and $A = P^{-1}DP$ say the same thing, namely, $A, B$ are similar, then why can't these equalities be equivalent? – 2017-02-12
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    These are the same only if $P=P^{-1}$. Otherwise, $P^{-1}AP=D$ is only equivalent to $A=PDP^{-1}$. You should try some examples by hand if you're not convinced, but this is pretty basic. To get from the first equation to the second, you basically multiply both sides by $P$ on the left and by $P^{-1}$ on the right (remember that matrix multiplication is not commutative, so the side matters!). – 2017-02-12
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    I see. What do these equalities say in words, though? $P^{-1}AP = D$ means $D$ is similar to $A$ and $A$ is similar to $D$. $P^{-1}DP = A$ means $A$ is similar to $D$ and $D$ is similar to $A$. Is that correct? – 2017-02-12
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    Yes. Notice, however, that just because each equation implies $A$ and $D$ are similar, they do not imply that the similarity matrix is the same in each direction. – 2017-02-12
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    Great. Thanks @ Fimpellizieri. – 2017-02-12
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Your chain of reasoning breaks down right away. $A=P^{-1}DP$ does not imply $A=PDP^{-1}$. You make a similar error in one of your earlier questions.

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    @ amd, I am using the following proposition: https://s23.postimg.org/mowa6tvdn/333.png I just let $m = 1.$ – 2017-02-12