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11 votes

Kind of submultiplicativity of the Frobenius norm: $\|AB\|_F \leq \|A\|_2\|B\|_F$?

A simpler, more direct proof that requires no SVD: let $Y_j$ be the $j$th column of $Y$ and $Z_j$ that of $Z=XY$. Then, $$\|Z\|_F^2 = \sum_j \|Z_j\|_2^2 = \sum_j \|XY_j\|_2^2 \leq \sum_j \|X\|_2^2\|Y_ …
Federico Poloni's user avatar
3 votes

Can a perturbation of a matrix product always be represented as product of perturbations of ...

In numerical analysis lingo, you are more or less asking if matrix multiplication is backward stable. The answer seems to be no: see Section 3.5 of Higham, Accuracy and stability of numerical algorith …
Federico Poloni's user avatar
2 votes
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Null Space Perturbations

What follows is just a trivial manipulation, but I do not think you can get better bounds than that for a generic perturbation. … leq \lVert \Delta A \rVert / \lVert A \rVert$ (up to first order in $\lVert \Delta A \rVert$, otherwise you have to take into account the additional term in the denominator depending on how large your perturbation
Federico Poloni's user avatar