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Reduce the cost of opnorm of B_k #133

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MohamedLaghdafHABIBOULLAH opened this issue Mar 1, 2024 · 8 comments
Open

Reduce the cost of opnorm of B_k #133

MohamedLaghdafHABIBOULLAH opened this issue Mar 1, 2024 · 8 comments

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@MohamedLaghdafHABIBOULLAH
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MohamedLaghdafHABIBOULLAH commented Mar 1, 2024

"It could be interesting to try other norms that are less expensive than the spectral norm, particularly the trace norm (the sum of eigenvalues when $B_k>0$)

@MohamedLaghdafHABIBOULLAH MohamedLaghdafHABIBOULLAH changed the title Reduce the cost of opnorm of $$B_k$$ Reduce the cost of opnorm of $B_k$ Mar 1, 2024
@MohamedLaghdafHABIBOULLAH MohamedLaghdafHABIBOULLAH changed the title Reduce the cost of opnorm of $B_k$ Reduce the cost of opnorm of B_k Mar 1, 2024
@dpo
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dpo commented Mar 1, 2024

The Frobenius norm is also interesting because $\Vert A \Vert_2 \leq \Vert A \Vert_F \leq \Vert A \Vert_*$, and it’s cheap to compute.

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dpo commented Sep 9, 2024

I think the convergence theory continues to hold if we compute any $\beta_k = \Theta(\Vert B_k \Vert)$, but we have to decide on how to compute $\beta_k$, especially for quasi-Newton operators.

@MohamedLaghdafHABIBOULLAH
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If we use the Frobenius norm, then we solve the allocation issue with opnrom.

@dpo
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dpo commented Sep 9, 2024

But we don’t have it for quasi-Newton operators.

@MohamedLaghdafHABIBOULLAH
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Ok, I thought that we have norm for LinearOperators of all type...

@MohamedLaghdafHABIBOULLAH
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Otherwise, the trace norm if B_k is positive, and I believe that we can easily extract the diagonal

@dpo
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dpo commented Sep 10, 2024

For the trace norm, we need the singular values. It's even costlier.

@MohamedLaghdafHABIBOULLAH
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Nope, if B_k is symmetric positive, then the trace norm is just the sum of its diagonal which is a norm!

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