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.

1. File->properties->Initial view
2. Edit->preferences->Page display->Default layout and zoom
3. Edit->preferences->Accessibility->Override page display

Many people recommend the second way, but it does not work for me.

The first option works, but only for one single PDF.

The third one works for all PDF for me!!! I found it here

%% basics %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% define a variable
\def\length{3}

% coordinate
\coordinate (A)  at (0,0);
% relative coordinate: example
\draw ($(A)+(0,1)$) -- ($(A)+(1,0)$);

% arrow and text
\draw[->,thin] (A)--(A1) node[above] {$x$}; % the text x is beside the node A1
% arrow types
->
->>
-latex
-stealth
-angle 90 % looks like ->
% path and text
\path (1) edge[bend left=7,->,sloped] node[near start]{\contour{white}{\tiny -1}  }(2);



Suppose S is a real skew-symmetric matrix; H is a real symmetric matrix.

For a real vector x, we have $x^TSx=0$. But for a complex vector x, we do not have $x^*Sx=0$. Instead, we have

• $x^*Sx$ is imaginary since $(x^*Sx)^*=x^*S^*x=-x^*Sx$ (if the conjugate of a complex number has the opposite sign, then the complex number is imaginary)
• $x^*Hx$ is real since $(x^*Hx)^*=x^*Hx$ (if the conjugate of a complex number is itself, then it is real)

Here are two basic identities about the monotonicity of binormial coefficients

• ${n \choose k}\ge {n \choose k-1}$ for any $0\le k \le \lceil n/2\rceil$
In addition, ${n \choose \lceil n/2\rceil}={n \choose \lfloor n/2\rfloor}$
For proof, see the post in stackexchange by Mario
• ${n+1 \choose k}\ge {n \choose k}$ for any $0\le k\le n$
For proof, see the post in stackexchange