Eureka! I figured it out! It's quite silly because in fact I was so close.

So yes, my hypothesis is true.

Actually, Since $W_1=\{c_1x_1+c_2x_2+...+c_kx_k|c_i\in\mathbb{C}\}$ is obviously a subspace of $W_0=\mathbb{C}^n$, and we can easily deduce (same approach as the first part of my proof) that $Bx\in W_1$ as long as $x\in W_1$, we obtain that $B$ is $W_1$-invariant, or we can say $B$ is also a linear transformation on $W_1$, so that there are $k$ eigenvalues of $B$ in $W_1$.

Now, pick one of them, say $\lambda_B$. As we can clearly see it's the first part of my proof all over again. For any $C\in\mathscr{F}$, we can get a new subspace $W_2\subseteq W_1\subseteq W_0$, that $C$ is $W_2$-invariant, and go ahead we get $W_3\supseteq W_4\supseteq ...$

Note that, $W_0$ has a finite dimension $n$, and every time we have $1\le\dim{W_{i+1}}\le\dim{W_i}$ so there gotta be a subspace $$W_\infty=\bigcap_i{W_i}$$ with $\dim{W_\infty}\ge1$, that $\forall x\in W_\infty(x\ne 0)$ is an common eigenvector of all the matrices in $\mathscr{F}$.



As inspired by Matt.E's answer in that question, we know that $\forall A\in\mathscr{F}$, if $A$ has $k$ eigenvalues, then there are at least $k$ common eigenvectors in $\mathscr{F}$. Simply because we can do the same process for each of its eigenvalues.

How do I prove that in every commuting family there is a common eigenvector?
Hello I'm trying to do opposite stuff to this, because now it change from 2015/Sau/01 to 2015/01/01 I don't know how to change back to 2015/Sau/01, how can I do this with this script, because I need to keep locale language, please help me. 

var date = f.task_end_date.value.split("/");
var months = ['Sau', 'Vas', 'Kov', 'Bal', 'Geg', 'Bir','Lie', 'Rgp', 'Rgs', 'Spa', 'Lap', 'Grd'];
for(var j=0;j&lt;months.length;j++){
    if(date[1]==months[j]){
         date[1]=months.indexOf(months[j])+1;
     }                      
} 
if(date[1]&lt;10){
    date[1]='0'+date[1];
}                        
var formattedDate = date[0]+date[1]+date[2];

