For two square matrices $A$ and $B$, if $\left | A - B \right |^2_F$ is a small scalar number then can we assume $A \approx B$?

Question

When researching a way to evaluate if two square matrices are equal or (very close to being equal) for a computer vision localization problem, I came across this Math Exchange post

Minimize the Frobenius norm of the difference of two matrices with respect to matrix: $\underset{B} {\mathrm{argmin}} \left\| A- B \right\|_F$

Following the accepted answer in that post, is it "mathematically" sound to conclude that $A \approx B$ if $\left | A - B \right |^2_F$ is a very small scalar number?

With best

Answer 1

This depends on what properties of the matrices are under consideration. E.g. invertibility is not stable under the Frobenius norm. Set $$ A=\begin{bmatrix} \varepsilon &0\\0&\varepsilon \end{bmatrix}\text{ and }B=\begin{bmatrix} 0&0\\0&0 \end{bmatrix} $$ then $$ |A-B|_F=\sqrt{\sum_{i=1}^n\sum_{j=1}^n |a_{ij}|^2}=\sqrt{2}\,\varepsilon . $$ The matrices are close by that measure for positive and small $\varepsilon $ but have fundamentally different properties; one is regular, and the other one eliminates all. If you are depending on algebraic properties as invertibility, then matrix norms aren't suited to measure a distance. However, they represent a good measure in case you are interested in whether $A$ and $B$ are topologically, or geometrically close.