Question

How to prove these two important properties of Frobenius norm :
1. For the identity matrix, ||I||F=Sqrt[n], whereas ||I||1=||I||infinity = 1.

Answer 1

\[For the identity matrix, \left| \right|I \left| \right|_{F}=\sqrt{n}, whereas \left| \right|I \left| \right|_{1}=\left| \right|I \left| \right|_{\infty}=1.\]

Answer 2

2. \[\left| \right|A \left| \right|_{F}^{2}=trace \left( A ^{T}A \right)\] where \[trace \left( A \right)\]is defined as the sum of the diagonal elements of A. That is, if \[A=\left( a _{ij} \right), then \] \[trace \left( A \right)=\sum_{i=1}^{n} a _{ii}\]

Answer 3

I'm not even sure what the first one means.

Answer 4

Doesn't say where \(F\) or \(n\) come from.

Answer 5

F stand for Frebonius norm and n is for \[\mathbb{R} ^{n \times n}\]

Answer 6

since I is identity matrix

Answer 7

What is \[
\|I\|_1
\]Supposed to mean?

Answer 8

its mean one norm form identity matrix.

Answer 9

Okay can you actually give some definitions?

Answer 10

You generally need a definition to prove properties.

Answer 11

Okay..

Answer 12

Given a matrix A and a vector norm \[\left| \right|.\left| \right|\] a non-negative number defined by \[\left| \right|A \left| \right|_{p}=\max_{x \neq 0}\frac{ \left| \right|A x \left| \right|_{p} }{ \left| \right|x \left| \right|_{p} }\]

Answer 13

Okay now what is the \(\|_p\)?

Answer 14

I only know about one norm

Answer 15

owh, its refer to p-norm where p is a real number greater than or equal to 1.