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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$\begingroup$ What does this have to do with matrix norm? You seem to be trying to compute the norm of a vector... $\endgroup$– Igor RivinCommented Jul 16, 2012 at 21:27
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$\begingroup$ sorry about confusion. I indeed meant vector norm $\endgroup$– arbitUser1401Commented Jul 16, 2012 at 21:29
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1 Answer
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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.
answered Jul 16, 2012 at 23:45
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$\begingroup$ 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 $\endgroup$ Commented Jul 17, 2012 at 1:45