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I understand that a householder matrix has eigenvalues of either 1 or -1, however I isn't clear to me why the determinant is -1. Clearly the determinant is equal to the product of the eigenvalues so there must be an odd number of eigenvalues that are -1. Any help is greatly appreciated.

user96723
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A Householder matrix has eigenvalues $\pm 1$. To see this, notice that if $u$ is orthogonal to the vector $v$ which was used to create the reflector, then $Pu = u$, i.e., $1$ is an eigenvalue of multiplicity $n-1$, since there are $n-1$ independent vectors orthogonal to $v$. Also, notice $Pv = -v$, and so $-1$ is an eigenvalue with multiplicity $1$. So, the determinant of a Householder matrix is $-1$, since the determinant of a matrix is the product of its eigenvalues.

  • Why is 1 an eigenvalue with multiplicity n-1? What I mean to say is why does it not have multiplicity n-3 for example? – user96723 Sep 25 '13 at 00:58
  • This is because there are $n-1$ independent vectors orthogonal to $v$. –  Sep 25 '13 at 01:01
  • Can you please explain why there are n-1 independent vectors orthogonal to v? Thank you – user96723 Sep 25 '13 at 01:24
  • because $\mathrm{dim}(\mathrm{span}{v})+\mathrm{dim}(\mathrm{span}{v}^{\perp})=n$ – Algebraic Pavel Sep 25 '13 at 07:30
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    to prove $dim(span{v}^{\perp})) = n-1$ we can use rank nullity theorem and define n by n matrix A having 0 in all elements except the last row is $v^T$ so now rank(A) = 1 (asssuming v is non-zero) so null(A) = n-1 but null space of A is the set of all vectors orthogonal to v so we are done. – KFkf Nov 15 '16 at 16:56
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Algebraically, this follows from the following two facts:

  1. Every Householder matrix can be written as $I-2vv^\ast$ for some unit vector $v$.
  2. For vectors $x$ and $y$, we have the formula $\det(I-xy^\ast)=1-y^\ast x$.
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