workable for Office 365

• clear formatting: control+shift+N

If you would like to change the languarate that the autocorrection is based on,

Review tab->Langurage->Set proofing languarage->Change the language from British to American and disable “Detect language automatically” (your computer is British)

You may also need to go to

File tab->Options->Language->Choose Editing language: from British to American English

str={};
for i=1:n
str=[str, strcat(‘idx=’,num2str(i))];
end
legend(str, ‘location’,’SouthWest’)

For any square matrix A, we have

$-\sigma_{n-k+1}(A) \le \lambda_k\left(\frac{A+A^T}{2}\right) \le \sigma_k(A)$

where $0\le\sigma_n\le\dots\le\sigma_1$ are the singular values, and $\lambda_n\le\dots\le \lambda_1$ are the eigenvalues.

Proof of the right half of the inequality:

See Topics in matrix analysis: page 151, Corollary 3.1.5

Proof of the left half of the inequality:

First, we have

$\lambda_k\left(\frac{-A-A^T}{2}\right)=-\lambda_{n-k+1}\left(\frac{A+A^T}{2}\right)$

To verify, it is easy to see $\lambda_1\left(\frac{-A-A^T}{2}\right)=-\lambda_{n}\left(\frac{A+A^T}{2}\right)$ and $\lambda_n\left(\frac{-A-A^T}{2}\right)=-\lambda_{1}\left(\frac{A+A^T}{2}\right)$.

Second, applying the right half of the inequality gives

$\lambda_k\left(\frac{A+A^T}{2}\right)=-\lambda_{n-k+1}\left(\frac{-A-A^T}{2}\right)\ge-\sigma_{n-k+1}(-A)=-\sigma_{n-k+1}(A)$

if you would like to achieve something like this:

1 2 3 4->1 2 3 4

-1 0->2 4

5 6->1 2

you should use

mod(i-1,n)+1

where n=4

It’s not that easy, but I finally fixed it.