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As said in the title, in which cases an invertible matrix is equal to the transpose? When is this: $ A^{-1} = A^{T} $ true?

If the matrix A is orthogonal?

Thank you!

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    Do you want to ask about invertible matrices that are equal to their transposes, or matrices $A$ such that $A^{-1}=A^T$? These are not the same thing.2012-06-10
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    Well I want to find out if this expression: $Q^{T}AQ$ is equal to this: $QAQ^{-1}$, where Q is an orthogonal matrix and A is a symmetric one. @ChrisEagle2012-06-10
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    @Chris It's equal to $Q^{-1}AQ$.2012-06-10
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    @talmid: So there is not a problem that the transpose is on the left at the 1st expression and the invertible is on the right at the 2nd? Thank you for your reply! :)2012-06-10
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    @Chris : Could you change your subject line if it's not what you mean? If you're asking for which matrices $A$ is $A^T$ the same as $A^{-1}$, then you're _not_ asking when an invertible matrix is equal to its transpose. That's a different question and has a different answer. Could you alter your subject line to reflect what you're trying to ask? It could say "In which cases is the inverse of a matrix equal to the transpose?". There's something to be said for being comprehensible.2012-06-10
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    @MichaelHardy: Is it ok now? Sorry for that, I am learning to use the English math terminology now.2012-06-10
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    @Chris Since $Q$ is orthogonal, $Q^T = Q^{-1}$ and then $Q^TAQ = Q^{-1}AQ$. If you want $Q^TAQ = QAQ^{-1}$, you have $Q^{-1}AQ=QAQ^{-1}$, and then $AQ^2 = Q^2A$, which I don't think is true in the general case. No problem.2012-06-10
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    @talmid: How did this: $AQ^2 = Q^2A$ occure? :S So it isn't correct to say that $Q^TAQ = QAQ^{-1}$ since Q is an orthogonal and A is symmetrix?2012-06-10
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    I will post a new question as this is not what I've asked at the title. Sorry for the invonvenience :S What I asked at the title was answered :)2012-06-10
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    @Chris Multiplying by $Q$ to the right and to the left.2012-06-11
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    Look OK now. ${}{}$2012-06-11

2 Answers 2

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If $A^{-1}=A^T$, then $A^TA=I$. This means that each column has unit length and is perpendicular to every other column. That means it is an orthonormal matrix.

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    at the risk of reviving a dodgy question, may I ask "why" the geometric interpretation of *orthogonal matrix* is equivalent to the algebraic definition you gave? I know the property, but I don't understand it.2013-12-27
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    @TrevorAlexander: Think of $A$ as an arrangement of $n$ columns (each $n$ elements tall). Then the $(i,j)$ element of $A^TA$ is the dot product of the $i^\text{th}$ and $j^\text{th}$ columns of $A$ since the $i^\text{th}$ row of $A^T$ is the $i^\text{th}$ column of $A$.2013-12-27
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    could you give me confidence that this is actually an "if and only if"? I mean that **both** directions hold: $A^{-1} = A^\top \Leftrightarrow A^\top A = I$2014-03-25
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    @MillaWell: $A^{-1}=A^T\implies A^TA=I$: Multiply both sides on the right by $A$. $A^TA=I\implies A^{-1}=A^T$: By definition.2014-03-25
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You're right. This is the definition of orthogonal matrix.