The norm or the singular values of the sum of identity matrix and a rank-$1$ matrix

Question

Let $A$ be an $N \times N$ rank-$1$ matrix. I am interested in finding the norm or the maximum singular value of $(A-cI)$ where $I$ is $N \times N$ identity matrix and $c>0$ is a scalar constant.

Answer 1

If $N=1$ then $\|A-cI\| = |A-c|$.

Assume that $N>1$.

Note that since $\ker A $ is non trivial have $(A-cI)x = -cx$ for some non zero $x$, and so $\|A-cI\| \ge |c|$.

If $A$ is rank one it can be written as $A=u v^T$ for two vectors $u,v$. Without loss of generality we can take $\|u\| = 1$.

If $u,v$ are colinear, then $A=k u u^T$, for some $k$, and the eigenvalues of the symmetric $A$ are $k,0$, hence $\|A-cI\| = \max(|c|,|k-c|)$.

Assume that $u,v$ are not colinear (equivalently, they are linearly independent).

Now assume that $N=2$. The $N>2$ case will be dealt with subsequently.

Let $B=(uv^T -c I) (v^Tu-cI) = v v^T +c^2I -c(u v^T + v u^T)$. We want to compute $\sqrt{\lambda_\max(B)}$.

Note that $\lambda_\max(B) = c^2+\lambda_\max(C)$, where $C=v v^T -c(u v^T + v u^T)$.

In the basis $u,v$, the matrix $C$ has the representation $\begin{bmatrix} -c u^Tv & -c \|v\|^2 \\ u^T v - c & \|v\|^2-c u^T v \end{bmatrix} = \begin{bmatrix} 0 & -c \|v\|^2 \\ u^T v - c & \|v\|^2 \end{bmatrix} - c u^TvI$. The eigenvalues of the last matrix are ${1 \over 2} (\|v\|^2 \pm \sqrt{\|v\|^4+4 \|v\|^2c(c-u^Tv)})$, and hence the eigenvalues of $B$ (which are non negative) are ${1 \over 2} (\|v\|^2 \pm \sqrt{\|v\|^4+4 \|v\|^2c(c-u^Tv)}) +c (c-u^Tv)$. Hence $\|A-cI\| = \sqrt{{1 \over 2} (\|v\|^2 + \sqrt{\|v\|^4+4 \|v\|^2c(c-u^Tv)}) + c(c-u^Tv)}$.

If $N>2$, then $B$ has additional eigenvalues at $c^2$, hence the formula remains the same, since we know that the norm of $A-cI$ restricted to the subspace $\operatorname{sp}\{u,v\}$ is no less than $|c|$.