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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?

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  • $\begingroup$ How does $ad=bc$ follow from eq2? $\endgroup$
    – Divide1918
    Jun 29, 2020 at 15:27
  • $\begingroup$ Also imo this might be difficult to generalize to higher dimension. $\endgroup$
    – Divide1918
    Jun 29, 2020 at 15:31
  • $\begingroup$ $A$ is a reflection, so it need not have $+1$ as an eigenvalue. For example take $n=1$, $A=-1$, so $A-1\ne0$. $\endgroup$ Jun 29, 2020 at 15:32
  • $\begingroup$ Are you sure it's not $|A+I|=0$? $\endgroup$ Jun 29, 2020 at 15:33
  • $\begingroup$ What is an even orthogonal matrix? $\endgroup$
    – Divide1918
    Jun 29, 2020 at 15:34

1 Answer 1

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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.

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

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