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sorry I don't know how to post Latex here I have to implement in Matlab the following formula and I'd like to know if the expression given here really corresponds to the minimum indicated, and why:

enter image description here

I found several formula errors in the previous equation of the same publication, but here I'm unable to see if this is correct. So, is this correct, and how does one prove it? thanks

PS: the non-bold letters are matrices, I being the identity and the bold letters being column vectors, lambda is a scalar value

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What algorithm are you trying to implement, and where did this formula come from? –  J. M. Aug 21 '11 at 13:13
    
I'm not sure I phrased my subject correctly, I am just asking if the second line is actually the same equation as the first. It is related to this publication: ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=991022&tag=1 but it hardly matters as I'm simply interested in this formula without any context –  lezebulon Aug 21 '11 at 13:16
    
"second line is actually the same equation as the first." - no, what's being said is that the value of $\mathbf f$ that minimizes the expression after the $\arg\min$ is given by the (solution of the) second expression. I'm assuming here $\|\cdot\|$ is the 2-norm; with that, $\|\mathbf v\|^2=\mathbf v\cdot\mathbf v$... take the gradient of the expression of the expression within $\arg\min$ and see if you can rearrange it into something that looks like the second equation. –  J. M. Aug 21 '11 at 16:30

1 Answer 1

up vote 1 down vote accepted

Under the assumption the $f,g,\mu$ are finite-dimensional vectors in $\mathbb{R}^n$, $M$ is a square real matrix, $V$ is a tall real matrix and $\lambda$ is a scalar (Lagrange multiplier) and the norm is the Euclidean norm:

I think it is correct (though I wasn't really careful) What you can do is to write out the norms explicitly ($\| x \|^2 = x^Tx$) and then take the derivative with respect to $f$ (vector derivatives are not complicated, check this page) and then set it to zero. From convexity, that would give the unique global minimizer $f^*$.

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