I have SPD matrix A and two vectors z and b.
Is there exist a norm where I can calculate $A^{1/2}bz$ without having to calculate $A^{1/2}b$ explicitly ?
I have SPD matrix A and two vectors z and b. Is there exist a norm where I can calculate $A^{1/2}bz$ without having to calculate $A^{1/2}b$ explicitly ? 


I'm assuming that by $\ A^{1/2}bz \$, you're referring to the 2norm 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}bz\$. $\ A^{1/2}bz \_{2}^{2}=(A^{1/2}bz)^{T}(A^{1/2}bz)$ $\ A^{1/2}bz \_{2}^{2}=b^{T}Ab2b^{T}A^{1/2}z + z^{T}z$ $\ A^{1/2}bz \_{2}=\sqrt{b^{T}Ab2b^{T}A^{1/2}z + z^{T}z}$ Note that in many cases, the Cholesky factorization of $A$ can be used in place of the symmetric matrix square root. 

