Eigenvalues of Householder matrix 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?
 A: The characteristic polynomial of $H$ is
$$
p(\lambda)=|\lambda I-H|=(\lambda-1)^n+Tr(2uu^T)(\lambda-1)^{n-1}
$$
for $uu^T$ is a Rank-$1$ matrix and all principal minors above $2$ are $0$.
Since $Tr(2uu^T)=2Tr(u^Tu)=2$
$$p(\lambda)=(\lambda+1)(\lambda-1)^{n-1}$$
Or eigenvalues of $H$ are $±1$.
A: 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 $1$.
A: Just compute $Hu$ and $Hv$ where $v$ is orthogonal to $u$. And remember that eigen spaces of different eigenvalues are orthogonal (for symmetric matrices).
