Why eigenvalues of an orthogonal matrix made with QR decomposition include -1?

Question

I want to make a real orthogonal matrix whose eigenvalues don't include -1. However, eigenvalues of a matrix $Q \in \mathbb{R}^{n\times n}$ ($n$ is even number) made with QR decomposition of a random matrix always include -1.

e.g.

n = 10
Q,R = qr(rand(n,n))

Question.

1. Why eigenvalues of $Q$ include -1? Does this proceed from the algorithm of QR decomposition?
2. Is there any idea for get an orthogonal matrix whose eigenvalues doesn't include -1.

---

Edit 1

I noticed that "if a block diagonal matrix consists of $2\times 2$ appropriate rotation matrices, this matrix satisfies the condition".

i.e.

$$Q := \begin{bmatrix} \cos(\theta_1)&\sin(-\theta_1)&&&\\ \sin(\theta_1)&\cos(\theta_1)&&&\\ &&\cos(\theta_2)&\sin(-\theta_2)&\\ &&\sin(\theta_2)&\cos(\theta_2)&\\ &&&&\ddots \end{bmatrix}$$

But I want to make more complicated $Q$ as a test matrix.

---

Edit 2

An answer for the question 1

Let $Q\in \mathbb{R}^{2m \times 2m}$ be an orthogonal matrix.

(i) Determinant of a householder matrix is $-1$ ($\because$ Wikipedia)

(ii) If $Q$ is made with QR decomposition, $$ \begin{align} \det(Q) &= \det(H_1)\det(H_2)\cdots\det(H_{2m-1})\\ & = -1 \end{align} $$

(iii) $\det{Q}=-1 \Rightarrow -1 \text{ is an eigenvalue of } Q.$

Answer 1

I assume you mean that $A$ is a randomly generated matrix and $A=QR$ is its QR-decomposition.

1. The (real) eigenvalues of an orthogonal matrix can only be $\pm 1$. If you are randomly generating $A$, then you are effectively randomly generating $Q$ as well. Since $Q$ is $n\times n$, it will have $n$ eigenvalues (not counting multiplicity), and effectively, each eigenvalue has a 50-50 chance of being 1 or -1. Thus the chance of $Q$ having at least one negative eigenvalue is $1-\frac{1}{2^n}$, which for $n$ large (i.e. n=10) is pretty small.

2. If you replace any orthogonal matrix $Q$ with $Q^2$ then all the eigenvalues will be $+1$ (since if $v$ is an eigenvector of $Q$ corresponding to $-1$, then $Q^2v = (-1)(-1) v = v$ so all the eigenvalues of $Q^2$ are 1). Of course, this will change the matrix $A$, but it might be suitable for your purposes.