What is the determinant of the following Householder matrix
$$ I - 2\hat{x}\hat{x}^\top $$
where $\hat{x}$ is a normalized vector, i.e., $\|\hat{x}\| = 1$?
I have been told this is always $-1$ somehow but can't find proof of it anywhere.
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Possible duplicate: https://math.stackexchange.com/questions/219731/determinant-of-rank-one-perturbations-of-invertible-matrices – user326159 Sep 06 '21 at 10:23
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@user326159 thank you, that's very helpful! – Sep 06 '21 at 10:36
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1@Physics_Student From the information in the linked question, you should now be able to answer your own question. – Rodrigo de Azevedo Sep 06 '21 at 10:39
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The eigenvalues of the matrix $\hat{x} \hat{x}^T $ are $(n-1)$ zeros and one eigenvalue of $1$. Therefore, the eigenvalues of $(I - 2 \hat{x} \hat{x}^T) $ are $(n-1)$ $1's$ and one eigenvalue of $(-1)$. Since the determinant is equal to the product of the eigenvalues, then the determinant will be $(-1)$.
Hosam Hajeer
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