# If $A$ is an orthogonal matrix with $|A|=-1$, show that $|I-A|=0$

Let $$A$$ be an $$n \times n$$ orthogonal matrix where $$A$$ is of even order with $$|A|=-1.$$ Show that, $$|I-A|=0,$$ where $$I$$ denotes the $$n \times n$$ identity matrix.

My approach

$$A \cdot A^{\top}=I$$

$$|A| \cdot\left|A^{\top}\right|=1 \quad$$ or $$\quad\left|A^{\top}\right|=-1.....(2)$$

let $$A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$$

$$L = I-A$$

$$L=\left[\begin{array}{cc}1-a & -b \\ -c & 1-d\end{array}\right]; \quad 2 a d=2 b c$$(from eq2)

$$|L|=(1-a)(1-d) - bc$$

$$=-(a+d)$$

What to do next? Am I going wrong?

• How does $ad=bc$ follow from eq2? Jun 29, 2020 at 15:27
• Also imo this might be difficult to generalize to higher dimension. Jun 29, 2020 at 15:31
• $A$ is a reflection, so it need not have $+1$ as an eigenvalue. For example take $n=1$, $A=-1$, so $A-1\ne0$. Jun 29, 2020 at 15:32
• Are you sure it's not $|A+I|=0$? Jun 29, 2020 at 15:33
• What is an even orthogonal matrix? Jun 29, 2020 at 15:34

Since $$A$$ is orthogonal, rewrite $$\det(I-A)$$ as follows: \begin{align}\det(I-A)&=\det(A^TA-A)\\&=\det(A)\det(A^T-I)\\&=\det(A)\det((A^T-I)^T)\\&=-\det(A-I)\end{align} It follows that $$\det(I-A)=(-1)^{n+1}\det(I-A)$$ As $$n$$ is even, we get $$\boxed{\det(I-A)=0}$$

as desired.

• @ VIVID how do u able to take whole transpose?
– user791682
Jun 29, 2020 at 15:43
• @AmartyaRoy $\det(M)=\det(M^T)$ Jun 29, 2020 at 15:44
• one more doubt , how det((A^T−I)^T)=det(A−I)?
– user791682
Jun 29, 2020 at 15:54
• @AmartyaRoy, Using the facts $(A+B)^T=A^T+B^T$ and $(A^T)^T=A$, one obtains $$(A^T-I)^T=(A^T)^T-I^T=A-I$$ Jun 29, 2020 at 15:58
• thanks a lot for your response! I don't understand why I didn't notice this fundamental property.so silly of me :(
– user791682
Jun 29, 2020 at 16:04