Understanding the Norm Inequality ||Av|| ≤ ||A||||v||

[c0der]

Hi,

With the following norm inequality:

||Av|| ≤ ||A||||v|| implies ||A|| = sup_v [ ||Av||/||v|| ]

I understand that sup is the upper bound of a set B, or least upper bound if B is a subset of A, where the upper bounds are elements of both B and A.

Is this saying that the norm of A is the maximum of the set ||Av||/||v||, where there are multiple vectors v being considered?

 

[HallsofIvy]

I am puzzled by your use of words here. Of course, you mean "A" and "B" to be sets of numbers. But if a set of numbers has an upper bound, then it has an infinite number of upper bounds so I am puzzled by "sup is the upper bound of a set B, or least upper bound if B is a subset of A". The "sup(B)" is the "least upper bound" in any case.

And I am puzzled by "where the upper bounds are elements of both B and A." In general upper bounds of sets are NOT in the sets. At most the least upper bound of set A can be in A, in which case it is the maximum of set A. Generally we use the term "sup" specifically to handle those sets that do NOT have a maximum.

 

[Mark44]

Hi,

With the following norm inequality:

||Av|| ≤ ||A||||v|| implies ||A|| = sup_v [ ||Av||/||v|| ]

I understand that sup is the upper bound of a set B, or least upper bound if B is a subset of A, where the upper bounds are elements of both B and A.

In your formula above, A is not a set - it's a matrix.

Is this saying that the norm of A is the maximum of the set ||Av||/||v||, where there are multiple vectors v being considered?

Yes.