# KpNormDual

 Other toolboxes required kpNormDual Computes the dual of the (k,p)-norm of a vector or matrix none kpNormKyFanNormSchattenNormTraceNorm Norms yes (convex)

kpNormDual is a function that computes the dual of the (k,p)-norm of a vector or matrix. More explicitly, the (k,p)-norm of a vector $x = (x_1,x_2,\ldots,x_n)$ is $\|x\|_{(k,p)} := \left(\sum_{j=1}^k \big|x_i^\downarrow\big|^p \right)^{1/p},$ where $(x_1^\downarrow,x_2^\downarrow,\ldots,x_n^\downarrow)$ is a rearrangement of the vector $x$ with the property that $|x_1^\downarrow| \geq |x_2^\downarrow| \geq \cdots \geq |x_n^\downarrow|$. Similarly, the (k,p)-norm of a matrix is the (k,p)-norm of its vector of singular values. This function computes the dual of this norm, which is fairly complicated and was derived in[1]. This function works with both full and sparse vectors and matrices.

## Syntax

• NRM = kpNormDual(X,K,P)

## Argument descriptions

• X: A vector or matrix to have its norm computed.
• K: A positive integer.
• P: A real number ≥ 1, or Inf.

## Examples

### A simple 4-by-4 example

The (k,p)-norm of a matrix when k = 1 is simply the operator norm. The dual of the operator norm is the trace norm, so when k = 1 this function just returns the trace norm (regardless of p):

>> X = [1 1 1 1;1 2 3 4;1 4 9 16;1 8 27 64];
>> [kpNormDual(X,1,1), TraceNorm(X)]

ans =

77.0015   77.0015

Similarly, if K = min(size(X)) and P = 2 then kpNorm(X,K,P) is the Frobenius norm, which is its own dual. Thus kpNormDual(X,K,2) decreases from the trace norm of X to its Frobenius norm as K increases:

>> [kpNormDual(X,1,2), TraceNorm(X)]

ans =

77.0015   77.0015

>> kpNormDual(X,2,2)

ans =

72.6903

>> kpNormDual(X,3,2)

ans =

72.6505

>> [kpNormDual(X,4,2), norm(X,'fro')]

ans =

72.6498   72.6498