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
- $\exists k$ such that the eigenvalue $\lambda_k<0$ with $k \neq 1$
- 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
- rank $2$
- traceless
I try to use the following algorithm:
- Keep $i=1$ and $i=k$ and remove other terms,
- Let $$\bar{\lambda}= \frac{|\lambda_1|+|\lambda_k|}{2}=\frac{\sigma_1+\sigma_k}{2}$$
- 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!