# In which cases is the inverse matrix equal to the transpose?

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. –  Chris Eagle Jun 10 '12 at 23:10
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. @ChrisEagle –  Chris Jun 10 '12 at 23:13
@Chris It's equal to $Q^{-1}AQ$. –  talmid Jun 10 '12 at 23:16
@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! :) –  Chris Jun 10 '12 at 23:26
@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. –  Michael Hardy Jun 10 '12 at 23:27

## 2 Answers

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 is an orthogonal 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. –  Trevor Alexander Dec 27 '13 at 8:22
@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 $A$. –  robjohn Dec 27 '13 at 10:03
could you give me confidence that this is actually an "if and only if"? I mean that both directions hold: $A^{-1} = A^\top \Leftrightarrow A^\top A = I$ –  Milla Well Mar 25 '14 at 17:20
@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. –  robjohn Mar 25 '14 at 18:40

You're right. This is the definition of orthogonal matrix.

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