Matrix approximation by eigenvalues approximation

We know that given a symmetric $$n\times n$$ matrix $$M$$ with rank $$r$$, by SVD, we have $$M = \sum_{i=1}^r\sigma_i u_iu_i^T,$$ with $$\sigma_i >0$$ for $$i = 1,\dots,r$$.

We also know that the rank $$k approximation of $$M$$, $$\hat{M}$$, in the sense that $$\|M-\hat{M}\|$$ is minimized is $$\hat{M} = \sum_{i=1}^k \sigma_iu_iu_i^T.$$

My question is, can we further approximate $$\hat{M}$$, $$\tilde{M}$$, by adjusting singular/eigenvlaues of $$M$$?

For example, replace $$\sigma_1$$ by $$\tilde{\sigma}_1$$ such that for some matrix norm $$\|\cdot\|_?$$ $$\|\hat{M}-\tilde{M}\|_?$$ is minimized?

Vigorously speaking, no. Since you mentioned that $$M$$ is a symmetric matrix, the SVD you're stating simplifies into the EVD format $$M = VDV^T$$. Note that the 2-tuple $$(V,D)$$ are uniquely given by the eigenspace $$\mathrm{eig}(M) = \{(\lambda_i, v_i): Mv_i = \lambda_iv_i\}$$ of the matrix $$M$$. This means the matrix is exactly represented by the eigenvalues and the eigenvectors. The rank-$$r$$ approximation of $$M$$ follows from the heuristic that supposing $$\lambda_1\geq\lambda_2\geq\dots\geq\lambda_n$$ and if $$\lambda_1 \gg \lambda_{r + 1}$$ we can ignore $$\lambda_{r+1} \dots \lambda_{n}$$. So this time the approximation $$\hat{M}$$ is uniquely identified by the 2-tuples $$\{(\lambda_i, v_i)\}_{i = 1}^r$$. This gives $$\hat{M} = V_r\mathrm{diag}({\lambda_1}, \dots, {\lambda_r})V_r^T$$. So you cannot further approximate a rank-$$r$$ matrix that gives a better" approximation to $$M$$.
In practice, however, your numerical tools rounds the exact value of $$\lambda_i$$'s to a certain data space, e.g. 16-bit float or whatever. So one way you may try is to round $$\lambda_i$$'s up to $$\bar{\lambda}_i$$'s, say, 2 decimal places and reconstruct the rank-$$r$$ approximated matrix $$\bar{M} = V_r \mathrm{diag}(\bar{\lambda}_1, \dots, \bar{\lambda}_r) V_r^T$$ to approximate the matrix $$\hat{M}$$ using fewer space, while using the same rank.