If $A$ is a square matrix of order 3 with $\det(A)=1$ and $AA^\top=I$, prove that $\det(A-I)=0$.

Question

If $A$ is a square matrix of order 3 with $\det(A)=1$ and $AA^\top=I$, prove that $\det(A-I)=0$.

I tried everything I know but couldn't get to the proof.

Answer 1

I'm sure it can be proven by matrix calculus, but here's a semi-geometric proof:

We have $A A^T=I$, so $A$ is an orthogonal matrix.

$A$ is of order 3, and because its determinant is +1, it must be the matrix of a rotation in 3D space. Therefore it can be put into the form $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos(\theta) & -\sin(\theta)\\ 0 & \sin(\theta) &\cos(\theta) \end{pmatrix} $$ Therefore, $\det(A-I)=0$

Answer 2

$det(A-I)=det(A-AA^\top)=det(A(I-A^\top))=det(I-A^\top)=det(I-A)=(-1)^3det(A-I) \implies det(A-I)=0$

Answer 3

$A - I = A - AA^T$

$\Rightarrow det ( A- I) = det ( A - AA^T ) $

$\Rightarrow det ( A- I) = det (A (I^T - A^T))$

$\Rightarrow$ det (A- I) = det (A)•det$(I^T - A^T)$

$\Rightarrow det ( A- I)$ = 1• (-) det $( A - I )^T$

$\Rightarrow det ( A- I) = (-) det (A - I)$

$\Rightarrow det ( A- I) = 0$