positive semidefiniteness: a psd matrix substracted by another rank 1 psd matrix Given that $A$ is a positive semidefinite matrix, $x$ is a vector, $\lambda_0 \in [0, +\infty) $ is a real non-negative number. I want to know the answer to the following optimization problem.
$$
\arg \min_{\lambda} |\lambda- \lambda_0| \\
s.t. \;\;  A-\lambda xx^T \ge 0
$$
Note $A-\lambda xx^T \ge 0$ means that $A-\lambda xx^T$ is a positive semidefinite matrix.
 A: The required minimiser is $\min(\lambda_0,\lambda_1)$, where $\lambda_1 = \max\{\lambda\in\mathbb{R}: A-\lambda xx^T\succeq0\}$. So, the problem boils down to finding $\lambda_1$.
Let $Q$ be a real orthogonal matrix whose first column is $x/\|x\|$ and let $B=Q^TAQ$. Then the constraint $A-\lambda xx^T\succeq0$ is equivalent to $B-\lambda\|x\|^2E_{11}\succeq0$, where $E_{11}$ is the matrix with a $1$ at the $(1,1)$-th entry and zeroes elsewhere. Since $B$ is positive semidefinite, this constraint can be further transformed (by orthogonal similarity) into the form of
$$
\pmatrix{a-\lambda\|x\|^2 & v^T & u^T\\ v&D&0\\ u&0&0} \succeq 0
$$
where $a=x^TAx/\|x\|^2$ and $D$ is a positive diagonal matrix. Note that $u$ is necessarily zero because $B$ is positive semidefinite. Therefore, the above constraint is equivalent to
$$
M=\pmatrix{a-\lambda\|x\|^2 & v^T\\ v&D} \succeq 0.
$$
As $D$ is positive definite, $M$ is positive semidefinite if and only if $\det(M)\ge0$. Using Schur complement, we get $\det(M)=\det(D)(a-\lambda\|x\|^2-v^TD^{-1}v)$. Therefore
$$\lambda_1 = \frac1{\|x\|^2} (a - v^TD^{-1}v).$$
A: By Schur's Complement, $A-\lambda xx^T\succeq0$ is equivalent to
$ \pmatrix{\lambda^{-1}&x^T\\ x& A} \ge 0 $.
$ \pmatrix{\lambda^{-1}&x^T\\ x& A} $ can be diagonalized as
$ \pmatrix{\lambda^{-1}-x^TA^{-1}x  &0\\ 0& A} $.
$ \pmatrix{\lambda^{-1}-x^TA^{-1}x  &0\\ 0& A} \ge0$ iff $\lambda^{-1}-x^TA^{-1}x \ge0$ and $A \ge0$.
Now it is clear that $\lambda^{-1}-x^TA^{-1}x \ge0$ or 
$\lambda \le (x^TA^{-1}x)^{-1}$ 
should be the answer to this problem.
