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Why in Jacobi eigenvalues algorithm is is important to minimize $||B-A||_F^2$?

Where $B = J^T A J$.

"Matrix Computations" by Colub and Van Loan tells it at 8.4.2. Also you can find it here Jacobi Methods at the beginning of page 3.

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    Hint: let $A^{(m)} = G_{m-1}A^{(m-1)}G_{m-1}^T$ be the $m$-th matrix computed by the Jacobi method, and let $\lambda_1 \geq \dots \geq \lambda_n$ be the eigenvalues of $A$ and $\tilde a_{11}^{(m)} \geq \dots \geq \tilde a_{nn}^{(m)}$ is a reordering of diagonal elements of $A$, then $\| \lambda_i - \tilde a_{ii}^{(m)} \| \leq \| A^{(m)} - B\|_F.$ So minimzing $\|B-A\|_F$ approximates the eigenvalues of $A$. – 2017-01-17
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    @TheWaveLad If $A^{(\infty)} = \mbox{diag}(\lambda_{1'}, \dots, \lambda_{n'})$, then why $||A^{(\infty)}-B||_F\le ||A^{(m)}-B||_F.$ I know that $||A^{(\infty)}||_F = ||A^{(m)}||_F = ||B||_F.$ Maybe you could elaborate it to an answer? – 2017-01-17
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    One can define the "outer norm" $N(A) := \|A\|_F^2 - \sum_{i=1}^n |a_{ii}|^2$. Then $\|A^{(m)}-B\|_F = \sqrt{N(A^{(m)})}$. It can be shown that $N(A^{(m)}) = N(A^{(m+1)}) + 2|a_{ij}^{(m)}|$, so $N(A^{(m+1)}) \leq N(A^{(m)})$ and $\|A^{(\infty)} - B\|_F \leq \| A^{(m)} - B\|_F$ – 2017-01-17
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    Also note that $\lim_{m \to \infty} N(A^{(m)}) = 0$. So with increasing $m$ we approximate the eigenvalues. If we want to do that, we have to minimize $\|B-A\|_F^2$. I'd like to formulate this as an answer, but I'm not sure what's left to say. – 2017-01-17
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    @TheWaveLad i was able to show that $N(A^{(m)}) = N(A^{(m+1)}) + 2|a_{ij}^{(m)}|^2,$ but still can't show that $||A^{(\infty)}-B||_F \le ||A^{(m)}-B||_F.$ That is the last piece i need. That would be pretty much an answer. – 2017-01-18
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    Is $A^{(\infty)} = \lim_{m \to \infty} A^{(m)}$? Then $\|A^{(\infty)} - B\|_F \leq \|A^{(m)} - B\|_F$ would follow from the fact that $N(A^{(k)})$ is decreasing and approaching zero for $k \to \infty$. Or did I make a mistake? – 2017-01-18
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    @TheWaveLad Well, that's right, $N(A^{(k)})$ is decreasing, but $\mbox{diag}(A^{(\infty)})$ is increasing. I don't know how to show which part changes faster and contribute more. – 2017-01-18

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