I'm reading the article about inverse kinematics problem and there is an optimization problem: need to find such $\Delta\theta$ which minimizes following quantity:
$||J\Delta\theta - \vec{e}||^{2}+\lambda^{2}||\Delta\theta||^{2}$
Where $J$ is $[n \times n]$ matrix, $\Delta\theta$ and $\vec{e}$ are $[n \times 1]$ column vectors and $\lambda$ is a scalar constant. Title claims that this problem is equals to minimizing:
$\| \begin{pmatrix} J\\ \lambda I \end{pmatrix} \Delta\theta- \begin{pmatrix} \vec{e}\\ 0 \end{pmatrix} \|$
However I can not see why this has to be truth. If we expand braces we will get:
$\| \begin{pmatrix} J\Delta\theta-\vec{e}\\ \lambda\Delta\theta \end{pmatrix} \|$ where vector of size $[n\times1]$ is placed in each row.
Here problem begins. I thought that matrix can be written as a vector with $2n$ elements(just expand vectors it consists from) :
$ \begin{pmatrix} m_{1}\\ m_{2}\\ ...\\ m_{2n} \end{pmatrix} $
Then taking magnitude of it will give us: $ \sqrt{\sum\limits_{i=1}^{n}m_{i}^{2} + \sum\limits_{j=n+1}^{2n}m_{j}^{2}} $ where $\sum\limits_{i=1}^{n}m_{i}^{2}=\|J\Delta\theta-\vec{e}\|$ and $\sum\limits_{j=n+1}^{2n}m_{j}^{2} = \|\lambda\Delta\theta\|$.
Thus, expression can be written as: $\sqrt{\|J\Delta\theta-\vec{e}\| + \lambda\|\Delta\theta\|}$ which is far from being the first one. Actually I believe that transformed equation has to be squared and this is just a typo. However this way I am still missing squares around $\|J\Delta\theta-\vec{e}\|$ and $\lambda\|\Delta\theta\|$.
Where am I wrong? Intuitively I would say that I just can not expand vectors inside of matrix and have to take magnitude as a sum of squared magnitudes of its constituent vectors but why? I could not find any formula.
This is incorrect $\sum\limits_{i=1}^{n}m_{i}^{2}=\|J\Delta\theta-\vec{e}\|$ It is $\sum\limits_{i=1}^{n}m_{i}^{2}=\|J\Delta\theta-\vec{e}\|^2$
With this correction, the rest follows as in the article you are reading. You need to consider too that if you are minimizing a norm, a definite positive function of its arguments, you can minimize its square instead to get the same solutions ($x,y\gt 0 \implies (x\gt y\iff x^2\gt y^2)$
$$\sqrt{\|J\Delta\theta-\vec{e}\|^2 + \lambda^2\|\Delta\theta\|^2}\to \|J\Delta\theta-\vec{e}\|^2 + \lambda\|\Delta\theta\|^2$$