In which cases is the inverse of a matrix equal to its transpose, that is, when do we have $A^{-1} = A^{T}$? Is it when $A$ is orthogonal?

  • $\begingroup$ 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 :) $\endgroup$ – Chris Jun 10 '12 at 23:38
  • $\begingroup$ Look OK now. ${}{}$ $\endgroup$ – Michael Hardy Jun 11 '12 at 3:02

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

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


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