Show that $PQ - I_n$ is singular when $Q - I_n$ is not

Let $$Q$$ be an $$n \times n$$ real orthogonal matrix and $$u \in \mathbf{R}^n$$ be a unit vector. Let $$P = I_n - 2uu^\top$$ be the Householder reflection. Show that if $$Q - I_n$$ is not singular then $$PQ - I_n$$ is singular.

• An idea... Not sure it is a good one, but may worth a try. Use the canonical form for $Q$ and then work separately in the two dimensional spaces of the $R_i$ by dealing first with the case $n=2$. – mathcounterexamples.net Dec 9 '19 at 9:59

2 Answers

When $$n$$ is odd, $$Q$$ has a fixed point whenever it has determinant $$1$$. When $$n$$ is even, $$Q$$ has a fixed point whenever it has determinant $$-1$$. These facts can be seen by considering the eigenvalues of $$Q$$: there are always $$n$$ of them, they are closed (as a multiset) under conjugation, and they must multiply to the determinant. It follows that, when $$Q$$ does not have a fixed point, multiplying it by any negative-determinant orthogonal matrix produces a matrix with a fixed point.

• You should clarify that a matrix $M$ has a fixed point if and only if $M-I$ is singular – Ben Grossmann Dec 9 '19 at 10:51
• Your answer is correct, but not easily readable to an audience that is just learning linear algebra – Ben Grossmann Dec 9 '19 at 10:55

This answer isn't as neat as Christopher Gadzinski's, but it gives an explicit non-trivial solution to $$(PQ-I)v=0$$. Note that \begin{aligned} (I-Q)^{-1}+(I-Q^T)^{-1} &=(I-Q)^{-1}\left[(I-Q^T)+(I-Q)\right](I-Q^T)^{-1}\\ &=(I-Q)^{-1}\left[(I-Q)(I-Q^T)\right](I-Q^T)^{-1}=I. \end{aligned} Since the LHS is the double of the symmetric part of $$(I-Q)^{-1}$$, the above implies that $$2(I-Q)^{-1}=I+K$$ for some skew-symmetric matrix $$K$$. Hence $$1-2u^T(I-Q)^{-1}u=0$$, \begin{aligned} -P(PQ-I)(I-Q)^{-1}u &=(P-Q)(I-Q)^{-1}u\\ &=(I-Q-2uu^T)(I-Q)^{-1}u\\ &=\left[1-2u^T(I-Q)^{-1}u\right]u=0 \end{aligned} and $$PQ-I$$ is singular.