Show $\xi = \frac{(y - Mx)x^\ast}{x^\ast x}$ is the solution of the following

$$ (M + \xi)x = y $$

with the minimum $\Vert \xi\Vert_2$, where $M \in \mathbb{C}^{m \times n}, x \neq 0 \in \mathbb{C}^n, y \in \mathbb{C}$.


We have the form as follows:

$$ M + \xi = \frac{M x^\ast x}{x^\ast x} + \frac{(y - Mx)x^\ast}{x^\ast x} $$

I have noticed the form's similarity with the operator 2-norm:

$$ \sup_{x\neq0} \frac{\Vert Ax \Vert_2}{\Vert x \Vert_2} = \sup_{x\neq0} \frac{x^\ast A^\ast A x}{x^\ast x} $$

But here I cannot proceed. Any help will be appreciated.


From $\xi x = y - Mx$ we have $$ \|\xi\|_2\geq\frac{\|\xi x\|_2}{\|x\|_2}=\frac{\|y-Mx\|_2}{\|x\|_2}. $$ It means that whatever $\xi$ such that $(M+\xi)x=y$ we have, its 2-norm cannot be smaller than $\|y-Mx\|_2/\|x\|_2$.

Now just verify we get equality above with $\xi_\star=(y-Mx)x^*/\|x\|_2^2$.

Note that this solution is also unique. Any other solution $\xi$ can be expressed as $$ \xi=\xi_\star+\eta X_\perp^*, $$ where the columns of $X_\perp\in\mathbb{C}^{n\times (n-1)}$ are linearly independent and orthogonal to $x$ and $\eta\in\mathbb{C}^{m\times (n-1)}$ is a "parameter" matrix. If we also assume that the columns of $X_\perp$ are orthonormal then $$ \xi=\left[\frac{y-Mx}{\|x\|_2},\eta\right]\left[\frac{x}{\|x\|_2},X_\perp\right]^* $$ Since the matrix on the right side is unitary, we have $$ \|\xi\|_2=\left\|\left[\frac{y-Mx}{\|x\|_2},\eta\right]\right\|_2>\frac{\|y-Mx\|_2}{\|x\|_2}=\|\xi_\star\| $$ if $\eta\neq 0$.


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