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If $A$ is orthogonally diagonalisable, then $D = Q^TAQ.$ Also, $D = Q^TAQ \to A^m = QD^mQ^T$ where $m \in \mathbb Z.$

Since similarity of matrices is an equivalence relation, $D = Q^TAQ \to A = Q^TDQ.$ Thus $A = Q^TDQ = QDQ^T$.

Does that make sense? The reason I ask this question is because while proving another result, I used $Q^TDQ$ instead of $QDQ^T$ so if the equality above holds I don't have to change my proof.

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    No, the equality may not be true. – 2017-02-11
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    “Equivalence relation” and “equality” are not the same thing. – 2017-02-11
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    @ amd, if $A$ is similar to $D$, then $D$ is similar to $A$ by symmetry, right? By definition, if $A$ is similar to $D$, then $D = Q^TAQ$ and if $D$ is similar to $A$, then $A = Q^TDQ$. Does that make sense? – 2017-02-11
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    That is not true unless $Q^T=Q$ so you cannot use that in a general argument. – 2017-02-11
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    @ Test123, we always have $A = P^{-1}DP = PDP^{-1}$ where $P$ is some eigenvector matrix, right? Can't we, then, have $A = Q^{-1}DQ = QDQ^{-1}$ which implies(?) $A = Q^TDQ = QDQ^T$ since $Q^T = Q^{-1}$? – 2017-02-11

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