I hope my question is interesting enough to catch your attention!
Assume two matrices $A,B\in\mathbb{R}^{n\times n}$ where $\Vert A \Vert_F$>>$\Vert B \Vert_F$ and the operator $\Vert \bullet \Vert_F$ denotes the Frobenius norm of $\bullet$.
My question is as follows: under this hypothesis, is it possible to establish any result on the singular values of $(A+B)$? Like, for example, can I consider $B$ as a "perturbation" of the matrix $A$ and find some bounds on the singular values of $(A+B)$ in terms of $\Vert A\Vert$ and $\Vert B \Vert$?
Thanks in advance!!!
Best Answer
Suppose $ A, B \in \mathbb{R}^{n \times n}$ then we know the singular value decomposition exists for both of these. So, the following is true.
$$ A = U \Sigma V^{T} $$
with $ U,V, \Sigma \in \mathbb{R}^{n \times n} $ then we know $$\| A \|_{F} \leq \|U \| \|\Sigma\| \|V^{T} \| $$ but $U, V $ are orthogonal which means their norm is 1. So, we have $$ \|A \|_{F} \leq \|\Sigma\|_{F} $$ and $ \Sigma $ is diagonal matrix. The norm of $ \Sigma $ is $$ \| S \|_{F} = max_{i} |\sigma_{i}| $$ that is it is the maximum singular value which is the top entry incidently.
So, you have some bounds. From here then
$$ \|A+B\| \leq \| A \| +\|B\| $$ $$ \sigma_{A+B} \leq \sigma_{A} + \sigma_{B} $$