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Timeline for Euclidean norms of matrices

Current License: CC BY-SA 4.0

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Feb 10, 2022 at 14:49 answer added Simon timeline score: 3
Feb 10, 2022 at 11:03 comment added Geoff Robinson In general, if $A = [a_{ij}]$ is a (real or complex) $n \times n$ matrix, we have $\|A\|^{2} \geq \frac{\sum_{i,j} |a_{ij}|^{2}}{n}$ with equality if and only if $A$ is a scalar multiple of a unitary matrix, if that helps.
Feb 10, 2022 at 10:56 comment added Gro-Tsen Here is a specific example: $A = \begin{pmatrix}0&0&1\\0&1&1\\1&1&0\\\end{pmatrix}$ has $\|A\|^2 ≈ 3.25$ which is root of the irreducible polynomial $x^3 - 5x^2 + 6x - 1$ over $\mathbb{Q}$, so, algebraic of degree $3$. This suggests that it will be difficult to find an analogous formula for $3\times 3$ matrices.
Feb 10, 2022 at 10:46 comment added Gro-Tsen For what it's worth: noting that $\|A\|^2$ is the largest eigenvalue of $A^{\mathrm{t}}\,A$, so it is a root of the characteristic polynomial of the latter should give the above formula, implies that for $n\times n$ matrices with rational entries, $\|A\|^2$ is algebraic of degree $n$ at most, and suggests that we probably can't do much better in general than maybe a factor $\frac{1}{2}$ over this.
Feb 10, 2022 at 10:03 comment added coudy Yes, this is the operator norm associated to the euclidean norm. Edited.
Feb 10, 2022 at 10:03 history edited coudy CC BY-SA 4.0
edit for clarity.
Feb 9, 2022 at 23:45 comment added Nathaniel Johnston @LSpice - I believe that "induced norm" implies "operator norm".
Feb 9, 2022 at 23:41 comment added LSpice You specify a norm on $\mathbf R^2$, but not on $2\times2$ matrices. Do you mean to equip matrices with the operator norm?
Feb 9, 2022 at 23:41 history edited LSpice CC BY-SA 4.0
Capitalise title; other minor proofreading
Feb 9, 2022 at 23:33 history edited YCor
edited tags
Feb 9, 2022 at 22:27 history asked coudy CC BY-SA 4.0