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Let $textbf{A} in mathbb{C}^{n times n} $, such that $rank(textbf{A}) = r$ and the singular values of $textbf{A}$ be $sigma_{1} geq dots geq sigma_{r} > 0$.

Let $textbf{u}_j$ and $textbf{v}_j$ for $1 geq j geq r$ denote the left and right singular vectors of $textbf{A}$ respectively.

Now, suppose we have a matrix

$$textbf{B} = alphasum_{j=1}^{k}textbf{u}_j sigma_{j} textbf{v}_j^{*}$$

Where $alpha in mathbb{C}$ and $k < r$.

I want to find $|textbf{A} - textbf{B} |_{2}$, which is equivalent to the maximum singular value of $textbf{A} - textbf{B}$.

$textbf{Proof Attempt}$

The obvious move to me would be writing $textbf{A}$ as a sum of outer products.

$$textbf{A} = sum_{j=1}^{r}textbf{u}_j sigma_{j} textbf{v}_j^{*}$$

Rewriting $textbf{A} - textbf{B}$ gives

$$textbf{A} - textbf{B} = (1 - alpha)sum_{j=1}^{k}textbf{u}_j sigma_{j} textbf{v}_j^{*} + sum_{j=k+1}^{r}textbf{u}_j sigma_{j} textbf{v}_j^{*}$$

This is where I get stuck. Because $alpha$ is an arbitrary complex scalar, I don't know what can be said about how it affects the singular values in the first sum, since they need to be both real and non-negative.