2
$\begingroup$

I am currently studying a robotics paper, where a normalized frobenius norm of a matrix is calculated as part of determining the condition number of the corresponding linear system of equation:

enter image description here enter image description here

$\mathbf{W}$ is defined as $$\mathbf{W} = \frac{1}{n} \mathbf{I}$$ and inserting this definition into $\left|\left|\mathbf{J}\right|\right|$ results in $$\left|\left|\mathbf{J}\right|\right| = \sqrt{\text{tr}\left(\mathbf{J}\mathbf{W}\mathbf{J}^{T}\right)} =\sqrt{\text{tr}\left(\mathbf{J}\frac{1}{n}\mathbf{I}\mathbf{J}^{T}\right)} = \sqrt{\frac{1}{n}\text{tr}\left(\mathbf{J}\mathbf{I}\mathbf{J}^{T}\right)} = \sqrt{\frac{1}{n}\text{tr}\left(\mathbf{J}\mathbf{J}^{T}\right)} .$$

The normalization factor is therefore independet of $\mathbf{J}$. Therefore, what is the purpose of scaling $\left|\left|\mathbf{J}\right|\right|$ down by a factor proportional to its dimension?

In terms using $\kappa(\mathbf{J})$ to optimize a mechanism, this would mean that the respective value of each iteration would simply be scaled down by constant factor, and therefore not influence the results. Am I missing something here?

Thanks in advance!

$\endgroup$
2
  • 1
    $\begingroup$ Your equality does not hold; the 1/n belongs inside the root: $\sqrt{\mathrm{tr} \left ( \mathbf{J} \frac{1}{n} \mathbf{I} \mathbf{J}^T \right ) } = \sqrt{ \frac{1}{n} \mathrm{tr} \left ( \mathbf{J} \mathbf{I} \mathbf{J}^T \right ) }$. $\endgroup$
    – TimWescott
    Dec 21 '19 at 0:50
  • $\begingroup$ Yep, my mistake! $\endgroup$
    – Skydiver
    Dec 21 '19 at 1:20
2
$\begingroup$

It is because you are using the trace of the matrix, which is equal to the sum of the singular values. The ideal $\mathbf{JJ^T}$ has singular values all equal to 1, but for an $n \times n$ matrix this sum would be equal to $n$, not $1$, and the text is defining the ideal matrix norm as 1.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.