Let $C^*=C_X+\delta\otimes\delta$, where $(\delta\otimes\delta)h=\langle\delta,h\rangle\delta$. Suppose for contradiction that
$$
\langle\delta,\phi_l^*\rangle=0
\qquad(l=1,\ldots,K).
$$
Then $C^*\phi_l^*=C_X\phi_l^*$, so every leading eigenpair $(\lambda_l^*,\phi_l^*)$ is also an eigenpair of $C_X$. Since $C^*\geq C_X$, the <Courant–Fischer min-max principle> gives $\lambda_l^*\geq\lambda_l$. If $\lambda_l^*=\lambda_{j_l}$ in the strictly decreasing spectrum of $C_X$, this inequality implies $j_l\leq l$. Orthogonality and distinctness force
$$
\{j_1,\ldots,j_K\}=\{1,\ldots,K\}.
$$
Hence the first $K$ eigenvectors of $C^*$ span the same space as $\phi_1,\ldots,\phi_K$. Their assumed orthogonality to $\delta$ would imply $\langle\delta,\phi_k\rangle=0$ for every $k\leq K$, contradicting the hypothesis. Therefore some $l\leq K$ satisfies $\langle\delta,\phi_l^*\rangle\ne0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 225 / 2 / b / Solution

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For a compact positive self-adjoint operator $T$ with decreasing eigenvalues $\lambda_1\geq\lambda_2\geq\cdots$, the Courant–Fischer principle gives
$$
\lambda_k
=\min_{\dim V=k-1}\max_{\substack{x\perp V\\\lVert x\rVert=1}}
\langle Tx,x\rangle.
$$
In particular, $S\geq T$ in quadratic-form order implies $\lambda_k(S)\geq\lambda_k(T)$ for every $k$.


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 Incoming links: Courant–Fischer min-max principle