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I know that $A$ orthogonal $\Rightarrow$ |det($A$)| = 1. Now I need to prove or disprove the reversed statement:

$$ |\det(A)| = 1 \Rightarrow A \,\text{ is orthogonal} $$

This is what I'm currently trying:

$$ |\det(A)| = 1 \Rightarrow \det(A)^2 = 1 \Rightarrow \det(AA^t) = 1 $$

But I'm unsure whether this implies, that $AA^t = E_n$. Any help is welcome at this point. Maybe the statement isn't even true.

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    $\begingroup$ You won't be successful . Matrix satisfying equation $A^2=I$ has determinant which can be equal 1 but there are plenty such matrices which are not orthogonal. $\endgroup$
    – Widawensen
    Sep 28, 2017 at 14:19
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    $\begingroup$ Even diagonal matrix $A=\begin{bmatrix} 0.5 & 0 \\ 0 & 2 \end{bmatrix}$ has determinant equal $1$. $\endgroup$
    – Widawensen
    Sep 28, 2017 at 14:42

2 Answers 2

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It's not true:

$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$

has determinant $1$ but it's not orthogonal since the columns are not orthonormal.

Furthermore,

$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\cdot\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\cdot\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$

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    $\begingroup$ Thank you really much! I'll accept this answer as soon as possible :) $\endgroup$ Sep 28, 2017 at 14:19
  • $\begingroup$ @user1551 Thanks, corrected. $\endgroup$ Sep 28, 2017 at 14:50
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No. Take any matrix $A$ with determinant $d\not=0$ and divide the elements of the first row by $d$. Then the new matrix has determinant $1$. Now it should be easy to find a counterexample to your implication.

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