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I have a linear transformation $ A = I - 2vv^\top $ where $ v $ is n dimensional unit vector. I'm looking for a geometrical meaning for the transformation A. I can write $ A $ as $ A = (I - vv^\top ) - vv^\top $ so it is a projection onto the nullspace of $v$ minus the projection onto $v$. Any other geometrical insight I can gain for this transformation ?

Thanks!

Tomer
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    If I remember it right, it has something to do with the reflection w.r.t. the subspace that is orthogonal to $v$. – xbh Jul 18 '21 at 00:54
  • $A$ has $n-1$ eigenvalues $1$ and one eigenvalue $1-2|v|,$ the latter with eigenvectors $v.$ So $A$ fixes the space of vectors perpendicular to $v,$ and stretches in the direction $v.$ – Thomas Andrews Jul 18 '21 at 00:57
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    Try applying it to a vector $x = a v + w$, where $w$ is perpendicular to $v$. Does that help you see what it's doing? – eyeballfrog Jul 18 '21 at 00:57
  • since $v^{\top}v=1$ then $Av=Iv-2v=-v$ – janmarqz Jul 18 '21 at 00:59
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    These are known Householder matrices. Geometrically, they correspond to reflection. I recently gave an answer on the subject. – Theo Bendit Jul 18 '21 at 02:17
  • I think I get it - it is reflection over $ Null(vv^\top) $ in the direction of $v$. I defined a generic vector $ m = v_k + v_c $ where $ v_k $ is coming from the kernel and $ v_c $ from the column space of $ (vv^\top) $ and I get $ Am= A(v_k + v_c) = v_k - v_c $ – Tomer Jul 18 '21 at 04:25

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