Let $A \in \mathbb{R}^{n \times n}$ be a real matrix and $A'$ any matrix obtained from $A$ by replacing a single entry by $0$. It seems to hold experimentally that $\lVert A' \rVert_2 \leq \lVert A \rVert_2$, but I can't find a proof after a bunch of attempts. Any ideas?
Asked
Active
Viewed 93 times
3
-
What is $\lVert(a_{ij}){1\leqslant i,j\leqslant n}\rVert_2$? Is it $\sqrt{\sum{1\leqslant i,j\leqslant n}a_{ij}^{,2}}$? – José Carlos Santos Mar 27 '20 at 17:55
-
No, it is the $2$-norm $\lVert A \rVert_2 = \max_{\lVert u \rVert = 1} \lVert Au \rVert_2$ , or equivalently, the largest singular value of $A$ (see Matrix Norm article on Wikipedia for details). – smalldog Mar 27 '20 at 17:58
-
Interesting. Well, it certainly holds for diagonal matrices... – mostsquares Mar 27 '20 at 18:05
-
@chaos Right. I call it operator norm. – José Carlos Santos Mar 27 '20 at 18:10
-
if $A'$ is really the same as $A$ except a single component has been zero'd out, then $A'$ is not a submatrix of $A$ – user8675309 Mar 27 '20 at 20:23
-
Yes, you're right, the name sub-matrix is not really appropriate. – smalldog Mar 28 '20 at 08:46
1 Answers
2
This isn't true in general. Counterexample: $$ \left\|\pmatrix{1&1\\ 1&-1}\right\|=\sqrt{2}=1.4142 <1.6180=\frac{1+\sqrt{5}}{2}=\left\|\pmatrix{1&1\\ 1&0}\right\|. $$ However, if $A$ is entrywise nonnegative, it is true that $\|B\|_2\le\|A\|_2$ when $0\le B\le A$ entrywise. This is because $0\le B^TB\le A^TA$ implies that $\rho(B^TB)\le\rho(A^TA)$ (which follows from the more general fact that $X\le Y\Rightarrow\rho(X)\le\rho(Y)$ for nonnegative matrices $X$ and $Y$).
user1551
- 149,263
-
Thanks a lot. I made a mistake in my experiments and always had $A$ entrywise positive... How foolish. :) – smalldog Mar 28 '20 at 10:56