Cross-posted from my identical question on math.stackexchange:
Given a metrix $X$ and a vector $\vec{y}$, ordinary least squares (OLS) regression tries to find $\vec{c}$ such that $\left\| X \vec{c} - \vec{y} \right\|_2^2$ is minimal. (If we assume that $\left\| \vec{v}\right\|_2^2=\vec{v} \cdot \vec{v}$.)
Ridge regression tries to find $\vec{c}$ such that $\left\| X \vec{c} - \vec{y} \right\|_2^2 + \left\| \Gamma \vec{c} \right\|_2^2 $ is minimal.
However, I have an application where I need to minimize not the sum of squared errors, but the square root of this sum. Naturally, the square root is an increasing function, so this minimum will be at the same location, so the OLS regression will still give the same result. But will ridge regression?
On the one hand, I don't see how minimizing $\left\| X \vec{c} - \vec{y} \right\|_2^2 + \left\| \Gamma \vec{c} \right\|_2^2 $ will necessarily result in the same $\vec{c}$ as minimizing $\sqrt{ \left\| X \vec{c} - \vec{y} \right\|_2^2 } + \left\| \Gamma \vec{c} \right\|_2^2 $.
On the other hand, I've read (though never seen shown) that minimizing $\left\| X \vec{c} - \vec{y} \right\|_2^2 + \left\| \Gamma \vec{c} \right\|_2^2 $ (ridge regression) is identical to minimizing $\left\| X \vec{c} - \vec{y} \right\|_2^2$ under the constraint that $ \left\|\Gamma \vec{c}\right\|_2^2 < t$, where $t$ is some parameter. And if this is the case, then it should result in the same solution as minimizing $\sqrt{ \left\| X \vec{c} - \vec{y} \right\|_2^2}$ under the same constraint.