Show that characteristic equation of an orthogonal matrix is a reciprocal equation.


Given, $AA^t=I$, Let $f(\lambda)=|A-\lambda I_n|$ be the the associated characteristic polynomial of $A_{n\times n}$, $f(\lambda)=0$ being the characteristic equation. We have to show that $f(\lambda)=\pm \lambda^nf(1/\lambda)$. I am unable to show this. Please help me.

  • $\begingroup$ What is "a reciprocal equation"? "Reciprocal" of what? $\endgroup$ – DonAntonio May 22 '16 at 11:03
  • $\begingroup$ $f(\lambda)=0$ is reciprocal if $f(\lambda)=\pm \lambda^nf(1/\lambda)$ $\endgroup$ – user1942348 May 22 '16 at 11:05
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    $\begingroup$ And what is an ‘orthogonal matrix reciprocal equation’? $\endgroup$ – Bernard May 22 '16 at 11:07
  • $\begingroup$ @user1942348 Are you sure that's the definition? It doesn't seem to be very sound as what you are in fact asking is whether "$\;\lambda$ is a root $\;\implies\frac1\lambda\;$ is also a root. " is true... $\endgroup$ – DonAntonio May 22 '16 at 11:08
  • $\begingroup$ I have posted an answer based on the line of your query "We have to show that...". It is however true that the meaning of "reciprocal equation" in this context needs explanation. $\endgroup$ – Nephente May 22 '16 at 11:38


  1. I understand "reciprocal equation" as one that is unchanged when you replace $x$ by its reciprocal $1/x$. $$f(x) = 0 \Rightarrow f(1/x) = 0$$ The only definition I found is from googling around is http://www.thefreedictionary.com/Reciprocal+equation. It seems it's not a term (frequently) used in math literature.
  2. $1$ also means the unit matrix where appropriate.

From the multilinearity of the determinant it follows that for any $n\times n$ matrix $$ \det(\lambda A) = \lambda^n \det(A) $$

Let $A\in \mathrm{O}(n),\; AA^t=1$. It should be clear that $\det(A) = \det(A^t) = \pm 1$.

Now evaluate

$$ \lambda^n \det(A-\tfrac{1}{\lambda}I) = \det(\lambda A - I) = \det(\lambda A-AA^t) = \det(A) \det(\lambda - A^t)$$

Since $ \det(-M) = (-1)^n\det(M) $, and $\det(M) = \det(M^t)$ for any square matrix $M$, one finds

$$ \det(A) \det(\lambda - A^t) = (\pm 1)(-1)^n \det(A - \lambda) = \pm \det(A-\lambda)$$ as claimed.

  • $\begingroup$ Nice calculations. Thanks. Would you please explore "the meaning of "reciprocal equation" in this context needs explanation". Please explain "reciprocal equation" more. $\endgroup$ – user1942348 May 22 '16 at 12:34

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