TITLE: Eigenvalues of Householder matrix
QUESTION [5 upvotes]: What would be the eigenvalues for a Householder matrix defined as: $H = I - 2 u u^T$? Can someone explain it  to me intuitively or with a simple proof?

REPLY [9 votes]: Householder matrices are orthogonal and symmetric.  That is:
$$ (I - 2uu^T)^T = I - 2uu^T $$
$$ (I - 2uu^T)^2 = I - 4uu^T + 4u(u^Tu)u^T = I $$
Here $u^T u = 1$ since $u$ is a unit vector.
Eigenvalues of orthogonal matrices have absolute value $1$, since multiplication by an orthogonal matrix is an isometry (length preserving).
Since the Householder matrix $H = I - 2uu^T$ is real and symmetric, its eigenvalues are real.  The only real numbers with absolute value $1$ are $\pm 1$.
Since $Hu = u - 2u(u^T u) = -u$, there is at least one eigenvalue $-1$, and indeed the vectors $v$ perpendicular to $u$ will satisfy $Hv = v - 0u = v$.  So there is a complete basis of eigenvectors, one for the eigenvalue $-1$ and the rest for eigenvalue $1$.