# How Find $\cos{(\pi A)}$ if $A$ is Orthogonal matrix

let $A_{n\times n}$ is Orthogonal matrix,

Find the value $$\cos{(\pi A)}=?$$

and before I guess $$\cos{(\pi A)}=E-2A$$ Now this is wrong,and this problem relsut is what?

and we konw if $x\in Z$,then $$\cos{(\pi x)}=(-1)^x$$

and How find this value? Thank you

• What is $E$? The $n \times n$ identity matrix? Dec 13, 2013 at 5:56
• @RobertLewis,yes,Thank you
– user94270
Dec 13, 2013 at 5:57
• @ nanchangian: if $R$ here is the real number system, and then we take $x = 1/2$, so that indeed $x \in R$, then $\cos (\pi x) = 0$, but whatever $(-1)^{1/2}$ is, we certainly don't have $(-1)^{1/2} = 0$. So what's up? Dec 13, 2013 at 6:00
• Do you mean $A~{}$ special orthogonal? I.e. with $\det(A)=1$? Your equation fails with $A=-1\in O(\Bbb R)$. Dec 13, 2013 at 6:11
• Have you tried your equation with $A=e^{i\theta}\in O(\Bbb R^2)$, $\theta\in\Bbb R$? I don't think it can work. Dec 13, 2013 at 6:17

This identity looks bogus. An orthogonal matrix $A$ is normal and so can be diagonalized by its eigenvectors, and your identity reduces to $$\cos(\pi \lambda) = 1-2\lambda$$ for every eigenvalue $\lambda$ of $A$. Any complex number of modulus 1 can appear as an eigenvalue of a rotation matrix, and the above doesn't even hold for $\lambda=-1.$ So $$A=\left[\begin{array}{cc}-1 & 0\\0 & -1\end{array}\right]$$ provides an explicit counterexample.

• Even easier, let $A = [-1]$. Dec 13, 2013 at 6:28
• @RyanReich Yep, but I figured a special orthogonal matrix was the safest bet given the discussion above. Dec 13, 2013 at 6:30
• Oh, good point. Dec 13, 2013 at 6:30
• Thank you,@user7530,I have edit problem,
– user94270
Dec 13, 2013 at 6:34

I'm afraid the equation

$\cos (\pi A) = E - 2A, \tag{1}$

where $A$ is an orthogonal matrix, cannot hold, at least in the stated generality. To see a counterexample, consider the matrix

$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \tag{2}$

and look at $\cos (\pi J)$. Since

$J^2 = -E, \tag{3}$

it is easy to show that, exactly as in the case of $e^{i \theta} = \cos \theta + i \sin \theta$ where $\theta \in \Bbb R$, the real numbers (and in fact for $\theta$ complex as well), we have

$e^{\theta J} = (\cos \theta) E + (\sin \theta) J; \tag{4}$

in fact the demonstration of (4) may be had by means of the power series expansion of $e^{\theta J}$, which, by virtue of (3), looks frighteningly similar to that of $e^{i \theta}$; a more thorough discussion of this proof may be found in my answer to this question. In any event, having (4) in hand, and observing that $J$ is in fact orthogonal, since $J^T = -J$ so that $J^TJ = -JJ = -J^2 = E$, we see that

$e^{\pi J} = (\cos \pi) E= -E = \begin{bmatrix} -1 & 0\\ 0 & -1 \end{bmatrix}, \tag{5}$

but

$E - 2J = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}, \tag{6}$

yielding another specific counterexample, in addition to that of user7530 in his/her answer to this question. And of course the above is quite similar in spirit to the comment of Olivier Begasset, which provided some, but not all, of the inspiration for my efforts here.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!