 Solution

Let $C^{\ast }=C_X+\delta \otimes \delta$, where $(\delta \otimes \delta )h=⟨\delta ,h⟩\delta$. Suppose for contradiction thatThen $C^{\ast }{\varphi }_l^{\ast }=C_X{\varphi }_l^{\ast }$, so every leading eigenpair $({\lambda }_l^{\ast },{\varphi }_l^{\ast })$ is also an eigenpair of $C_X$. Since $C^{\ast }\ge C_X$, the Courant–Fischer min-max principle gives ${\lambda }_l^{\ast }\ge {\lambda }_l$. If ${\lambda }_l^{\ast }={\lambda }_{j_l}$ in the strictly decreasing spectrum of $C_X$, this inequality implies $j_l\le l$. Orthogonality and distinctness forceHence the first $K$ eigenvectors of $C^{\ast }$ span the same space as ${\varphi }_1,\dots ,{\varphi }_K$. Their assumed orthogonality to $\delta$ would imply $⟨\delta ,{\varphi }_k⟩=0$ for every $k\le K$, contradicting the hypothesis. Therefore some $l\le K$ satisfies $⟨\delta ,{\varphi }_l^{\ast }⟩\ne 0$.