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

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.

• 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. Sep 28, 2017 at 14:19
• Even diagonal matrix $A=\begin{bmatrix} 0.5 & 0 \\ 0 & 2 \end{bmatrix}$ has determinant equal $1$. Sep 28, 2017 at 14:42

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,

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

• Thank you really much! I'll accept this answer as soon as possible :) Sep 28, 2017 at 14:19
• @user1551 Thanks, corrected. Sep 28, 2017 at 14:50

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.