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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 ?

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What does this have to do with matrix norm? You seem to be trying to compute the norm of a vector... –  Igor Rivin Jul 16 '12 at 21:27
    
sorry about confusion. I indeed meant vector norm –  zimbra314 Jul 16 '12 at 21:29

1 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.

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that is the problem with 2 norm, that it entails computing $A^{1/2}z$ which yet remains the problem. That is why in computation of $A^{-1}b$ people use $A$ norm (e.g. conjugate gradient method) cause it overcomes the problem of computing $A^{-1}z$ where z remains an approximation of x –  zimbra314 Jul 17 '12 at 1:45

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