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

Question

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?

Also imo this might be difficult to generalize to higher dimension. — Divide1918, Jun 29 '20 at 15:31
$A$ is a reflection, so it need not have $+1$ as an eigenvalue. For example take $n=1$, $A=-1$, so $A-1\ne0$. — Chrystomath, Jun 29 '20 at 15:32
$A$ is similar to a block-diagonal matrix where the blocks are either scalar or $2 \times 2$ rotation matrices. Then show that the “actually diagonal” part cannot consist of only $-1$s. — Aphelli, Jun 29 '20 at 15:37
@Divide1918 I assumed even stood for matrix with an even order. — VIVID, Jun 29 '20 at 15:40

VIVID · Accepted Answer · 2021-02-16T17:18:00.510

5

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.

edited Feb 16 '21 at 17:18

answered Jun 29 '20 at 15:37

VIVID

11,604

@ VIVID how do u able to take whole transpose? – Jun 29 '20 at 15:43
@AmartyaRoy $\det(M)=\det(M^T)$ – VIVID Jun 29 '20 at 15:44
one more doubt , how det((A^T−I)^T)=det(A−I)? – Jun 29 '20 at 15:54
@AmartyaRoy, Using the facts $(A+B)^T=A^T+B^T$ and $(A^T)^T=A$, one obtains $$(A^T-I)^T=(A^T)^T-I^T=A-I$$ – VIVID Jun 29 '20 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 :( – Jun 29 '20 at 16:04

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

1 Answers1