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:

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$\mathbf{W}$ 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!

  • 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

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.


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