elementary matrices and eigenvalues

Question

consider the elementary matrix (which is $n*n$) $E=I-\alpha xy^T$ show that this matrix has at least n-1 eigenvalues that equal 1.

Answer 1

Calculate the matrix $xy^T$ for non zero vectors $x,y$. Observe its rank. From the rank deduce that it has $n-1$ eigenvalues as zero and the last one being $x^Ty$.

Answer 2

Let $x$ be an eigenvector of $E$ with eigenvalue $1$. So $Ex=x$. You have to find that there are at least $n-1$ linearly independent vectors satisfying $Ex=x$. This is equivalent to $(E-I)x=0 \rightarrow -\alpha x y^T = 0$. So the vectors satisfying $Ex=x$ are in $y^\perp$. Since $y^\perp$ is $n-1$ dimensional, we can find linearly independent $n-1$ eigenvectors. This is true for $\alpha$ is nonzero. The case $\alpha$ is zero is obvious.