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I have a symmetric matrix $M\in \mathcal{S}^n$ with rank $\mathbf{r}>2$. We can arrange its singular values by $$(\sigma_1=|\lambda_1|)\geq (\sigma_2=|\lambda_2|)\geq \dots \geq (\sigma_r=|\lambda_r|).$$

Suppose

  1. $\exists k$ such that the eigenvalue $\lambda_k<0$ with $k \neq 1$
  2. We want to find the rank 2 approximation.

By SVD, we have $$M = \sum_{i=1}^r |\lambda_i|\,\,\mathbf{u}_i\big(\text{sgn}(\lambda_i)\mathbf{u}_i\big)^T,$$ where $\text{sgn}$ represents the sign function.

We know the rank $2$ approximation $\tilde{M}$ of $M$ is $$\tilde{M}=\sum_{i=1}^2 |\lambda_i|\,\,\mathbf{u}_i\big(\text{sgn}(\lambda_i)\mathbf{u}_i\big)^T,$$ so $$\|M-\tilde{M}\|=|\lambda_3|=\sigma_3.$$


Now I want an approximation which is

  1. rank $2$
  2. traceless

I try to use the following algorithm:

  1. Keep $i=1$ and $i=k$ and remove other terms,
  2. Let $$\bar{\lambda}= \frac{|\lambda_1|+|\lambda_k|}{2}=\frac{\sigma_1+\sigma_k}{2}$$
  3. I come up with $$\overline{M} = \bar{\lambda}\mathbf{u}_1\mathbf{u}_1^T - \bar{\lambda}\mathbf{u}_k\mathbf{u}_k^T$$

Obviously, $\overline{M}$ is of rank $2$, symmetric and traceless.

Is this a good/proper approximation? Would you mind suggesting any better way/reference?

Thanks!

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  • $\begingroup$ What norm are you measuring the approximation in? I have an idea that might be optimal for the Frobenius norm, but I'm not sure about other norms like the operator/spectral norm. $\endgroup$ Feb 3, 2021 at 15:35
  • $\begingroup$ @NathanielJohnston According to the property of spectral norm ($l$-$2$ norm) of a matrix $X$, $\|X\|-\sigma_{\max}(X)$. If my memory is correct, the Frobenius norm will get the same answer for low rank approximation (I am not very sure). $\endgroup$ Feb 3, 2021 at 16:30
  • $\begingroup$ For usual low-rank approximation the Frobenius and spectral norms indeed get the same answer, but it's not obvious to me that you get the same answer for this variant of low-rank approximation. Do you prefer a particular norm, or would any norm be OK? $\endgroup$ Feb 3, 2021 at 17:24
  • $\begingroup$ @NathanielJohnston Any norm would be OK! This is just a different way to see the problem. $\endgroup$ Feb 4, 2021 at 0:48

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