On Orthogonal Matrices

Orthogonal Matrices can be considered as basis transformations in $\mathbb R^n$ . Therefore, considering the matrix as a conglomeration of unit orthogonal column vectors, we obtain for some orthogonal matrix Q,

$Q = (u_1, u_2, \dots, u_n )$

$Q^T Q \ = \ (u_1, u_2, \dots, u_n )^T (u_1, u_2, \dots, u_n )$

$(Q^T Q)_{ij} \ = \ \langleu_i, u_j\rangle \ = \ \delta_{ij}$

$Q^T Q \ = \ I.$

The final equation implies a nifty equation,

$Q^{-1} = Q^T.$

The following equation has interesting consequences. Suppose $Q$ does not have $1$ as an eigenvalue. Then, $Q - I$ must be invertible, and the following formula holds. Write

$(I - Q)^{-1} + (I - Q^T)^{-1} \ = \ (I - Q)^{-1} + (Q - I)^{-1}Q \ = \ (I - Q)^{-1} (I - Q) = I$

and thus

$(I - Q)^{-1} + (I - Q^T)^{-1} \ = \ I. \ \ \dots (1)$

We apply this formula to prove the following proposition.

Proposition. Let $P$ be a reflection in the form of $P = I - 2u^T u$ for some unit vector $u \in \mathbb R^n$ . Also, set $Q$ to be some orthogonal matrix that does not have $1$ as an eigenvalue. Then, $PQ$ must have an eigenvalue of $1$

proof. It suffices to show that there exists a vector $v$ that is nonzero and satisfies $PQ v = v$ , or equivalently $Qv = Pv$ . Rewriting this condition, we obtain

$(I - Q) v \ = \ 2 u u^T v \ = \ 2u \langle u, v \rangle.$

We wish to use equation (1), and a reasonable guess for $v$ would be $v = (I - Q)^{-1} u$ . Then, it suffices to show that

$u \ = \ 2 \langle u, v \rangle u$

or even $\langle u, v \rangle = \frac 1 2$ . This can be directly proved by left multiplying $u^T$ and right multiplying $u$ to equation (1) and comparing the resulting bilinear forms. □

An implication of this proposition is that a composition of a reflection and an absolute rotation must always fix some vector in $\mathbb R^n$

As a sidenote, I want to comment that the object $(I - A)^{-1}$ is an important matrix in linear algebra. If $A$ has a spectral radius strictly less than 1, the proposed inverse can be expanded by Neumann expansions. In the case where $Q$ is orthogonal, the spectral radius is exactly $1$ . Equation (1) is nice since it provides control over this inverse. Also, this object can be considered as an analogy of taking the power series of $f(x)$ around $1$ . If $x = 0$ does not give a fruitful expansion, repeat for $x = 1$ ; it is only reasonable to do so.

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