Traceless low rank approximation of a symmetric matrix by SVD

Question

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 $$\|M-\tilde{M}\|=|\lambda_3|=\sigma_3.$$

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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!