Let $A$ be a $5\times 5$ matrix with singular values $10,10^2,10^3,10^4,10^5$. Show that there cannot exist a non-invertible matrix $B$ such that $||A\vec x - B\vec x|| \le $1 with $||\vec x|| = 4$.
My attempt:
Assume that such a matrix does exist. Then $B$ has rank $\le 5$. We know that the following holds:
$$ ||A-A_k|| \le ||A-B|| $$
Where $A_k = U\Sigma_k V^T$ with $\Sigma_k$ having the $k$ largest singular values of $A$. This means that $||A_k||$ can be at most $10^4$ (using the operator norm which gives the largest singular value).
$$\begin{aligned} ||A\vec x - A_k\vec x|| &\le ||A\vec x - B\vec x|| \le 1 \newline ||A\vec x - A_k\vec x|| &= ||(A-A_k)\vec x|| \le ||A-A_k|| \cdot ||\vec x|| = 10^4 \cdot 4 \end{aligned} $$
So basically I get that $||A\vec x - A_k\vec x||$ is less than or equal to $1$ and $4 \cdot 10^4$, however I would ideally want to show that there is a contradiction somewhere, and this is not a contradiction.
If $B$ is singular then $Bx = 0$ for some $x$ with $\|x\|= 4$. Then $\|Ax-Bx\| = \|Ax\| \ge \sigma_\min (A) \|x\| = 40$.