most recent 30 from http://mathoverflow.net 2013-05-24T08:18:56Z http://mathoverflow.net/feeds/question/102389 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102389/ease-of-calculation-of-norm zimbra314 2012-07-16T21:22:11Z 2012-07-16T23:45:09Z

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

http://mathoverflow.net/questions/102389/ease-of-calculation-of-norm/102405#102405 Brian Borchers 2012-07-16T23:45:09Z 2012-07-16T23:45:09Z

<p>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 $A$. </p> <p>If you can precompute $A^{1/2}z$, then you can quickly compute $\|A^{1/2}b-z\|$.</p> <p>$\| A^{1/2}b-z \|_{2}^{2}=(A^{1/2}b-z)^{T}(A^{1/2}b-z)$</p> <p>$\| A^{1/2}b-z \|_{2}^{2}=b^{T}Ab-2b^{T}A^{1/2}z + z^{T}z$</p> <p>$\| A^{1/2}b-z \|_{2}=\sqrt{b^{T}Ab-2b^{T}A^{1/2}z + z^{T}z}$</p> <p>Note that in many cases, the Cholesky factorization of $A$ can be used in place of the symmetric matrix square root. </p>