Question

I have SPD matrix A and two vectors z and b.

Is there exist a norm where I can calculate $||A^{1/2}b-z||$ without having to calculate $A^{1/2}b$ explicitly ?

Answer 1

I'm assuming that by $\| A^{1/2}b-z \|$, you're referring to the 2-norm and that by $A^{1/2}$, you're referring to the unique symmetric matrix square root of $A$.

If you can precompute $A^{1/2}z$, then you can quickly compute $\|A^{1/2}b-z\|$.

$\| A^{1/2}b-z \|_{2}^{2}=(A^{1/2}b-z)^{T}(A^{1/2}b-z)$

$\| A^{1/2}b-z \|_{2}^{2}=b^{T}Ab-2b^{T}A^{1/2}z + z^{T}z$

$\| A^{1/2}b-z \|_{2}=\sqrt{b^{T}Ab-2b^{T}A^{1/2}z + z^{T}z}$

Note that in many cases, the Cholesky factorization of $A$ can be used in place of the symmetric matrix square root.