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I'd like to know whether every orthogonal matrix

$$ A \in \mathcal{O}_n(\mathbb{R})$$

can be written as a product of givens-rotations. I know that when we do QR-decomposition of matrix $A$ we get

$$ A = Q R $$

So my idea was to prove that $R$ must be the identity $I_n$, however I'm stuck at that. Can somebody give me a hint on how I could prove this?

Jean-Claude Arbaut
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msrd0
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    Orthogonal matrices could be reflections. – Doug M Nov 28 '18 at 18:01
  • https://www-old.math.gatech.edu/academic/courses/core/math2601/Web-notes/3num.pdf page 38 –  Nov 28 '18 at 18:06
  • You could use that rotations are orthogonal matrices and that the product of orthogonal matrices is orthogonal to prove that $R=Q^TA$ is an upper triangular Matrix which is orthogonal. Since, orthogonal matrices are normal matrices, we would conclude that $R$ is an diagonal matrix with 1 or -1 in the diagonal of $R$. That is, A is a composition of rotations or a composition of rotations with compositions of reflections, which is quite close of what you are asking. – R. W. Prado Aug 31 '21 at 11:00

2 Answers2

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Givens rotations are... rotations, they preserve orientation ($det(M)=+1$), however $\mathcal{O}_n(\mathbb{R})$ has two components, one component is rotations ($det(M)=+1$), the other is reflections ($det(M)=-1$).

Example of $M\in\mathcal{O}_2(\mathbb{R})$ that can not be written as a Givens rotation (its determinant is $-1$) $$ M=\left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right) $$ You can not find Givens rotation $G_\theta=\left(\begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\right)$ such that $M=G_\theta$

If you want to have a Givens rotation decomposition you must restrict yourself to $SO_n(\mathbb{R})$, the special orthogonal group, which is defined by $M^tM=I_d$ and $\det{M}=+1$.

Picaud Vincent
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    Thanks, I was trying to construct a matrix where it wouldn't work before attempting to prove, I must have missed that one – msrd0 Nov 28 '18 at 22:13
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If you write the matrix $A=QR$ then there are two ways you can perform this.

$$ A \underbrace{R_{1} R_{2} \cdots R_{n}}_{\hat{R}^{-1}} = \hat{Q} \tag{1} $$

or the following, which is what you want

$$ \underbrace{Q_{n} Q_{n-1} \cdots Q_{2}Q_{1}}_{{Q}^{*}}A = R \tag{2} $$

In $1$ we see that Gram-Schmidt applies a sequence of elementary triangular matrices $R_{k}$ on the right of $A$. For $2$ we see a sequence of elementary unitary matrices $Q_{k}$ on the left of $A$. The matrices $Q_{k}$ can be givens matrices.

$$ \underbrace{G_{n} G_{n-1} \cdots G_{2}G_{1}}_{{G}^{*}}A = R \tag{3} $$

All you need to prove is that $G_{k}$ is a unitary matrix. It is. There are notes provided here.