let A be a $ n \times n$ orthogonal matrix,where n is an even number with $|A|=-1\quad$. Show that $|A-I|=0$

So basically I have to show that 1 is an eigenvalue of A.

Here is how I proceeded:-

Since A is orthogonal $$\Rightarrow AA^T = I \Rightarrow AA^T -I =0 \Rightarrow AA^T -AA^{-1} =0 \Rightarrow A(A^T- A^{-1}) = 0$$ $$|A||(A^T- A^{-1})|=0 \Rightarrow|(A^T- A^{-1})|=0$$, since $|A|=-1$

If A has the eigenvalue $\alpha_1,\alpha_2,\alpha_3,......\alpha_n$

Then $A^T$ must also have the same eigenvalues $\alpha_1,\alpha_2,\alpha_3,......\alpha_n$

Then $A^{-1}$ has the eigenvalues of $ \frac{1}{\alpha_1},\frac{1}{\alpha_2},\frac{1}{\alpha_3},.....\frac{1}{\alpha_n}$

Then $A -A^{-1}$ have the eigenvalues in the form of $\alpha_i -\frac{1}{\alpha_i}$
**I am not sure whether this above mentioned statement is correct **

then $|A -A^{-1}|$ can be written as $$\prod_{i=1}^{i=n} \alpha_i -\frac{1}{\alpha_i} =0 \Rightarrow \alpha_t -\frac{1}{\alpha_t} =0\quad \text{,for some t}$$ $$\Rightarrow {\alpha_t}^2 =1 \Rightarrow {\alpha_t} =\pm 1 $$. This proves that some eigenvalue may take the value of +1, or -1.

I am not sure of how to proceed from here .Also I have a question about the purpose of even order of matrix A. Any other way to prove is welcome.

  • 1
    $\begingroup$ The conventional way to prove the statement is to note that $A-I=A(I-A^T)$. $\endgroup$ – user1551 May 1 '18 at 15:08
  • $\begingroup$ For some variants see here, here. I still think I have seen this handled on our site, so keeping my answer CW even though I could not find an exact duplicate. $\endgroup$ – Jyrki Lahtonen May 1 '18 at 15:26

Since $A$ is orthogonal, each of its eigenvalues is a complex number whith absolute value $1$. Since the eigenvalues are the roots of the characteristic polynomial of $A$, wich is a real polynomal, for each complex non-real eigenvalue $\lambda$, $\overline\lambda$ is also an eigenvalue and $\lambda.\overline\lambda=|\lambda|^2=1$. Therefore, $|A|$ is the product of the real eigenvalues of $A$, each of which is $\pm1$. Since the total number of eigenvalues is $n$, wich is even, and since the non-realeigenvalues come in pairs, there's an even number of real eigenvalues. Since their product is $-1$, they can't be all equal to $-1$.

  • $\begingroup$ Is there any proof of"Since A is orthogonal, each of its eigenvalues is a complex number whith absolute value 1?" $\endgroup$ – DRPR May 1 '18 at 15:19
  • $\begingroup$ @DRPR If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $\|Av\|^2=\langle Av,Av\rangle=\langle\lambda v,\lambda v\rangle=|\lambda|^2\|v\|^2$. But, on the other hand, $\|Av\|^2=\langle Av,Av\rangle=\langle v,A^TAv\rangle=\langle v,v\rangle=\|v\|^2$. Since $v\neq0$ and $|\lambda|^2\|v\|^2=\|v\|^2$, $|\lambda|=1$. $\endgroup$ – José Carlos Santos May 1 '18 at 15:25

Combine $$|A-I|=|A-AA^T|=|A|\cdot |I-A^T|=-|I-A^T|=-|I-A|$$ and $$|I-A|=(-1)^n|A-I|.$$

  • $\begingroup$ $-|I-A^T|=-|I-A|$ how? $\endgroup$ – DRPR May 1 '18 at 15:23
  • 1
    $\begingroup$ @DRPR For any square matrix $M$ we have $|M|=|M^T|$. Apply that to $M=I-A$. $\endgroup$ – Jyrki Lahtonen May 1 '18 at 15:27

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