Let $v\in\mathbb{R}^k$ with $v^Tv\neq 0$. Let $$P=I-2\frac{vv^T}{v^Tv}$$ where $I$ is the $k\times k\ $ identity matrix. Then which of the following statements is (are) true?

(a) $P^{-1}=I-P$

(b) $-1$ and $1$ are eigenvalues of $P$

(c) $P^{-1}=P$

(d) $(I+P)v= v$

I'd really appreciate it if someone could just run me through the process of checking each option while briefly touching upon the relevant properties.

  • $\begingroup$ Let $a$ be an arbitrary vector such that $v^Ta=0$. A straightforward calculation results in $$\eqalign{Pv&=-v\cr Pa&=a\cr}$$These are eigenvalue equations, and thus they answer (b) and (d). Multiplying these same vectors by $P^2$ will produce the answers to (a) and (c). $\endgroup$ – greg Jan 28 at 18:37
  • $\begingroup$ $P$ is an orthogonal matrix ( see en.wikipedia.org/wiki/Householder_transformation), whence options (b) and (c) are correct. $\endgroup$ – StubbornAtom Jan 28 at 20:56

For (a) and (c) you can check if $P P^{-1} = I$. For (d) try to verify by direct computation. For (b) try to check the eigenvalues for the eigenvectors $v$ and $w$ where $w$ is orthogonal to $v$, i.e. $w^T v = 0$.


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