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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.

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  • $\begingroup$ 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 $n=2$. $\endgroup$ – mathcounterexamples.net Dec 9 '19 at 9:59
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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.

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

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