Let $A\in\mathrm{SO}_{2k+1}\mathbb{R}$, then $A$ has $1$ as an eigenvalue. Why is this true? By using the characteristic polynomial, we see that $A$ must have some eigenvalue, but this tells us nothing about the particular value of that eigenvalue. I suspect the proof is trivial, but somehow, I can't think of it off the top of my head.
 A: Suppose that $\lambda$ is a real eigenvalue of $A$ and let $x$ be an eigenvector for $\lambda$ of norm $1$. Then $$1=\langle x,x\rangle=\langle Ax,Ax\rangle=\lambda^2\langle x,x\rangle$$ so that $\lambda$ is either $1$ or $-1$.
Now $A$ has a certain number of non-real eigenvalues, which come in conjugate pairs, a certain number of $1$s and a certain number of $-1$s. The product of all its eigenvalues is $1$, as that is the determinant of $A$. The product of the non-real eigenvalues is a positive real number. It follows that the multiplicity of $-1$ has to be even, for otherwise the product of all eigenvalues would be negative, and it isn't.
All this implies that the total multiplicity of all the eigenvalues different from $1$ is even. It follows that $1$ has odd multiplicity and therefore the multiplicity is at least equal to one.
A: Proffering a "holistic" approach.

Ok, so we know that 


*

*$\det A=1$,

*$AA^T=I$, and 

*$\det (-I)=(-1)^{2k+1}=-1$.


These imply that
$$
\begin{aligned}
\det(A-I)&=-\det(I-A)\\
&=-\det(AA^T-A)\\
&=-\det(A(A^T-I))\\
&=-\det(A)\det(A^T-I)\\
&=(-1)\det((A-I)^T)\\
&=-\det(A-I).
\end{aligned}
$$
This implies that $\det(A-I)=0$, and therefore $\lambda=1$ is an eigenvalue.
A: The number of real eigenvalues is odd and the product of all real eigenvalues yields the sign of the determinant (because the complex eigenvalues contribute with their positive absolute value to the sign). Thus not all real eigenvalues can be $-1$.
A: The characteristic polynomial $\chi_A$ of $A$ is a real polynomial of odd degree. Hence $\chi_A$ has real roots. Moreover the product of the roots is equal to $\det A=1$, and the modulus of each root is equal to $1$.
It implies that $1$ is a root of $\chi_A$ as the non real roots are conjugated.
A: Let $A\in SO_{2n+1}(\mathbb{R})$ such that $AA^\star=I_{2n+1}$.
$A^\star$ denotes the transposed matrix.
We know that $A$ is similar to
$$\mathrm{diag}\left(\underbrace{1,\cdots,1}_p,\underbrace{-1,\cdots,-1}_q,R(\theta_1),\cdots,R(\theta_k)\right)$$
and
$$R(\theta)=\pmatrix{\cos(\theta) & -\sin(\theta)\cr\sin(\theta) & \cos(\theta)}$$
This implies that $\det(A)=(-1)^q$, but we know that $\det(A)=1$, hence $q$ is even.
If $p$ was also even, the size of the matrix would be even : contradiction.
Hence $p$ is odd, and in particular $p\ge 1$.
As a conclusion, $1$ is an eigenvalue.
A: Every eigenvalue of an orthogonal matrix has a complex modulus 1, thus any real eigenvalue of $A$ is either $-1$ or $1$. If $A$ is in $\mathrm{SO}_{2k+1}\mathbb{R}$ then by definition the determinant of $A$ is $1$. 
If $\lambda$ is a non-real eigenvalue of $A$, then the complex conjugate $\bar{\lambda}$ is also an eigenvalue of $A$ because the characteristic polynomial of $A$ is a real polynomial. Thus $\lambda \bar{\lambda} = |\lambda|^2=1 > 0$. 
We know that the determinant of $A$ is the product of all eigenvalues of $A$. Thus, for all the non-real eigenvalues, they come in pairs, with their product all together being a positive number $1$. Since we have an odd number of eigenvalues, with their product being one, this means that $1$ has to be an eigenvalue, otherwise you would have an odd number of $-1$ with their product being $1$, which is impossible.
