Can we compare elements of two matrices based on their norm and singular values?

Question

Let $[\mathbf{U},\mathbf{\Sigma},\mathbf{V}]=svd(\mathbf{M})$ and define \begin{equation} \operatorname{Separability}(\mathbf{M}) = \frac{\sigma^{2}_1}{\sum_{i=1}^{n}{\sigma^2_i}}. \end{equation}

Now consider two unknown matrices $\mathbf{A}$ and $\mathbf{B}$. All we know about $\mathbf{A}$ and $\mathbf{B}$ is

1. Both $\mathbf{A}\in\mathbb{R}^{n\times n}$ and $\mathbf{B}\in\mathbb{R}^{n\times n}$.
2. $\|\mathbf{A}\|_2^2 \approx \|\mathbf{B}\|_2^2$
3. Let $\sigma^{\mathbf{A}}_{i}\geq0 \hspace{1mm}\forall \hspace{1mm}i=1,\cdots,n$ be singular-values of $\mathbf{A}$ and $\sigma^{\mathbf{B}}_{i}\geq0 \hspace{1mm}\forall \hspace{1mm}i=1,\cdots,n$ be singular-values of $\mathbf{B}$.

\begin{equation} \operatorname{Separability}(\mathbf{A}) \approx \operatorname{Separability}(\mathbf{B}) \end{equation}

With this above information, is it reasonable to say that $a_{i,j}\approx b_{i,j}$?

Answer 1

If $A=\begin{pmatrix}1&0\\0&x\end{pmatrix}$ and $B=\begin{pmatrix}x&0\\0&1\end{pmatrix}$ for some $x>0$, then all your conditions are met with equality for points 2 and 3. Your conclusion requires $x\approx1$, but that was not needed to satisfy your assumptions. Therefore what you want is not true as stated.