• Support PF! Buy your school textbooks, materials and every day products Here!

Proving Frobenius norm

  • Thread starter tinorina
  • Start date
  • #1
1
0

Homework Statement


Prove that the Frobenius norm is indeed a matrix norm.


Homework Equations


The definition of the the Frobenius norm is as follows:
||A||_F = sqrt{Ʃ(i=1..m)Ʃ(j=1..n)|A_ij|^2}


The Attempt at a Solution


I know that in order to prove that the Frobenius norm is indeed a matrix norm, it must satisfy the 3 properties of matrix norm, which are as follows:
1. f(A) >= 0, for all A in ℝ^(mxn) (f(A)=0 iff A=0)
2. f(A+B) <= f(A)+f(B), for all A, B in ℝ^(mxn)
3. f(αA) = |α|f(A), for all α in ℝ, A in ℝ^(mxn)

However, I'm not exactly sure how to go about proving each of the properties. Can someone please give me some hints? Thanks!
 

Answers and Replies

  • #2
133
4
Do you know any vectorspaces with similar norms?
Maybe you can relate the properties of those norms to this one!
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,705
1,722

Homework Statement


Prove that the Frobenius norm is indeed a matrix norm.


Homework Equations


The definition of the the Frobenius norm is as follows:
||A||_F = sqrt{Ʃ(i=1..m)Ʃ(j=1..n)|A_ij|^2}


The Attempt at a Solution


I know that in order to prove that the Frobenius norm is indeed a matrix norm, it must satisfy the 3 properties of matrix norm, which are as follows:
1. f(A) >= 0, for all A in ℝ^(mxn) (f(A)=0 iff A=0)
2. f(A+B) <= f(A)+f(B), for all A, B in ℝ^(mxn)
3. f(αA) = |α|f(A), for all α in ℝ, A in ℝ^(mxn)

However, I'm not exactly sure how to go about proving each of the properties. Can someone please give me some hints? Thanks!
So, what difficulties are you having proving property 1? Where is your problem proving property 3?

RGV
 

Related Threads for: Proving Frobenius norm

  • Last Post
Replies
0
Views
4K
  • Last Post
Replies
3
Views
937
  • Last Post
Replies
5
Views
669
  • Last Post
Replies
0
Views
1K
Replies
5
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
Top