# What does $||A|| = 1$ mean in the definition of QLSP?

In their 2017 paper, Childs et al. gave the definition of QLSP beginning with : "Let $A$ be an $N\times N$ Hermitian matrix with known condition numbers $\kappa$, $||A|| = 1$ and at most $d$ non-zero entries in any row or column..."

I initially thought that by $||A|| = 1$ they meant that QLSP requires spectral norm (largest eigenvalue) of $A$ to be $1$, which sounds reasonable to me as even the original HHL paper they needed the eigenvalues of $A$ to lie in between $1/\kappa$ and $1$.

But, the paper: Quantum linear systems algorithms: a primer seems to have interpreted is as "the determinant of $A$ needs to be $1$" in page 28 definition 6.

Which interpretation is correct and why? In case the latter is correct, I'm not sure why it is so. I don't see why the restriction that $\text{det}(A)$ needs to be $1$ even make sense. It (the unit determinant condition) doesn't even guarantee that the eigenvalues of $A$ will be less than or equal to $1$, which is necessary for HHL to work.

I presume you could rewrite conditions in terms of the determinant (you would have to alter the time step $t_0$) but it's not clear to me why you would want to. It's also worth noting that definition 8 (page 39) in that paper defines matrix inversion, putting limits on the eigenvalues of the matrix: bounding between some minimum value and 1. So they're also implicitly acknowledging that structure, and certainly not setting the determinant (product of eigenvalues) to 1.
• I think that was also a misinterpretation from them. But they say also that the unit determinant condition precludes that the matrix is not invertible. Maybe a simplification too unless $|| A || = 1$ precludes the non-inversibility case. – cnada Aug 21 '18 at 16:22