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

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    $\begingroup$ The usual QR decomposition algorithms use Housoholder reflectors. If you want to stay in SO(n), use Givens rotations in the QR decomposition. $\endgroup$ Nov 15, 2017 at 14:27
  • $\begingroup$ "Is there any idea for get an orthogonal matrix whose eigenvalues doesn't include -1." - in the even case, try multiplying your orthogonal matrix with the identity matrix, modified so that one randomly chosen entry is $-1$ instead of $1$. $\endgroup$ Nov 16, 2017 at 6:34

1 Answer 1

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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.

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    $\begingroup$ I think it is wrong that "The eigenvalues of an orthogonal matrix can only be ±1±1." A counter example is [0 -1; 1 0] (a rotation matrix whose angle is pi). By the way, the idea using Q^2 instead of Q is good for me. Thank you very much for your answer. $\endgroup$
    – Kgm1500
    Nov 15, 2017 at 11:47

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