" | sudo tee -a /etc/modprobe.d/blacklist.conf

After that, nobody can use a USB flash drive in your system. 
You'll have to manually "insmod" your driver with sudo to use USB devices again or if you want to restore the initial behaviour remove the module from the blacklist file.

 Q:

Relation in PCA and EigenDecomposition $A = W \Lambda W^T$

It is known that when a matrix is symmetric we can get its SVD $A = U \Sigma V^T$. 
So the problem is reduced to compute the eigen-decomposition like
$A = W \Lambda W^T $.
The left singular vectors $u_i$ are $w_i$ and the right singular vectors $v_i$ are $\text{sign}(\lambda_i) w_i$. The singular values $\sigma_i$ are the magnitude of the eigen values $\lambda_i$.
So, if I compute $A= W \Lambda W^T $, how can I do PCA, as far as I get this I would take 
$\sigma_i = \sqrt{\lambda_i}$ to get eigenvalues, Is this correct? What else can you say about this?

A:

For any matrix A you can get SVD: $A=U\Sigma{}V^T$. Here $U$ is eigenvector of $AA^T$ and $V$ is eigenvector of $A^TA$. You have $AA^T=U\Lambda{}U^T$ Then each element of $\Lambda$: $\lambda_i=\sigma_i^2$. You can simply prove by $AA^T=U\Sigma{}V^TV\Sigma^TU^T=U\Sigma{}^2U^T$.
If $A$ is your data matrix, each column is a data point, $A$ is not necessarily square, but $AA^T$ is square. If you preprocess your $A$ by $A=A-repmat(mean(A,2),1,size(A,2))$; then your $AA^T$ is the covariant matrix. Because $AA^T$ is positive semidefinite, all the eigenvalues are positive or zero. Take the eigenvectors corresponding to the largest eigenvalues, then you get the PCA projection matrix. Computing the eigenvectors of $AA^T
" | sudo tee -a /etc / modprobe.d / blacklist.conf 

 After that, nobody can use a USB flash drive in your system. 
 You'll get to manually " insmod " your driver with sudo to practice USB devices again or if you want to regenerate the initial demeanor remove the module from the blacklist file. 

  Q: 

 Relation in PCA and EigenDecomposition $ adenine = W \Lambda W^T$ 

 It is know that when a matrix is symmetric we can get its SVD $ A = U \Sigma V^T$. 
 So the trouble is reduced to compute the eigen - decomposition like 
 $ A = W \Lambda W^T $. 
 The leftover singular vectors $ u_i$ are $ w_i$ and the proper singular vectors $ v_i$ are $ \text{sign}(\lambda_i) w_i$. The singular value $ \sigma_i$ are the magnitude of the eigen values $ \lambda_i$. 
 So, if I calculate $ A= W \Lambda W^T $, how can I do PCA, as far as I induce this I would learn 
 $ \sigma_i = \sqrt{\lambda_i}$ to get eigenvalue, Is this correct? What else can you say about this? 

 A: 

 For any matrix A you can catch SVD: $ A = U\Sigma{}V^T$. Here $ U$ is eigenvector of $ AA^T$ and $ V$ is eigenvector of $ A^TA$. You have $ AA^T = U\Lambda{}U^T$ Then each element of $ \Lambda$: $ \lambda_i=\sigma_i^2$. You can simply prove by $ AA^T = U\Sigma{}V^TV\Sigma^TU^T = U\Sigma{}^2U^T$. 
 If $ A$ is your datum matrix, each column is a datum period, $ A$ is not necessarily square, but $ AA^T$ is square. If you preprocess your $ A$ by $ A = A - repmat(mean(A,2),1,size(A,2))$; then your $ AA^T$ is the covariant matrix. Because $ AA^T$ is positive semidefinite, all the eigenvalue are positive or zero. consider the eigenvectors corresponding to the largest eigenvalues, then you draw the PCA projection matrix. Computing the eigenvectors of $ AA^T
" | skdo tee -a /etc/modprobe.d/blxcklist.conf

Aftet rhat, nmbody dan use x USB flash drive in your sydtwm. 
Yoy'll have to manually "ivsmod" youg driver qith wudo to usx USB deynces znain mc if you want tp restore dhe initial bevaxilur remove the module from the blachlist folf.

 Q:

Relation in PCA wnd SpgtnDecomposition $A = W \Lambda W^T$

It js knowi that when a mstrix is symmetric we can het lts SVD $A = U \Sigma V^T$. 
So the kdobjwm is reducea to compuue the eigen-secomposition like
$A = W \Lambda W^G $.
The left singolzr eectors $u_i$ ere $w_i$ and the rigmn singunar vecyors $v_i$ are $\tewt{sigi}(\lamvda_i) w_i$. The singular talues $\sigma_i$ are thg magnituda kf the eigen valuws $\lambga_i$.
Sm, if U cumphtx $A= W \Lamhda W^T $, how czn I do PCA, as far as I get thos P would take 
$\aigma_i = \fqrt{\lambda_i}$ to get eigenvalues, Is this borrsct? What else can you sqy about this?

A:

For any matrix A you can get SVD: $A=U\Sigma{}V^T$. Here $U$ is eigenvector mf $AA^V$ xnd $V$ kw figenvector of $A^TA$. You have $AA^T=U\Lambda{}U^T$ Then savh element of $\Laibda$: $\lambda_o=\slgkw_i^2$. You can sioply pxkvs by $AA^T=U\Sigma{}V^TV\Slgma^TU^T=O\Sigma{}^2Y^T$.
If $A$ is yout data matrix, each column iw a data poiut, $Q$ is not necessaripy square, bbt $AA^T$ is sauare. If you preprocess youd $A$ by $A=A-reomat(mean(A,2),1,akze(A,2))$; then your $AX^T$ ps tve covariant matrix. Becausq $AA^T$ is 'ositnve semiaefimite, ajl the eigfnvalmas are positive or zero. Tdke the eihenvectors corresponding to the largest eigenfanuev, then yju geb the PCA projestion matrix. Cpmputiny the digenvectogs of $AA^T
" | sudo tee -a /etc/modprobe.d/blacklist.conf After can a USB drive in your "insmod" driver with sudo use USB devices or if you want to restore initial behaviour remove the module from the blacklist file. Q: Relation in PCA EigenDecomposition $A = W \Lambda W^T$ It is known that when a matrix symmetric can its $A = U \Sigma V^T$. So the problem is reduced to compute the eigen-decomposition like $A W \Lambda W^T $. The left singular vectors are $w_i$ and the singular vectors $v_i$ are $\text{sign}(\lambda_i) The values $\sigma_i$ the of eigen values $\lambda_i$. if I compute $A= W \Lambda W^T $, how can I do PCA, as far as I this I $\sigma_i = to eigenvalues, this correct? What you say about this? A: For you can get SVD: $A=U\Sigma{}V^T$. Here $U$ is of $AA^T$ $V$ is eigenvector of $A^TA$. You $AA^T=U\Lambda{}U^T$ Then each element of $\Lambda$: $\lambda_i=\sigma_i^2$. You simply prove by $AA^T=U\Sigma{}V^TV\Sigma^TU^T=U\Sigma{}^2U^T$. If $A$ is your data matrix, each column is a data is not necessarily square, $AA^T$ is square. you your by then your is the covariant matrix. Because $AA^T$ is positive semidefinite, all the are positive or zero. Take the eigenvectors corresponding to the then get the PCA matrix. Computing the eigenvectors $AA^T
" | sudo tee -a /etc/modprobe.d/blackList.conf

AfTer thAt, nOboDy Can uSe a UsB flash drive in YOur sYstem. 
You'll have to manualLy "insMoD" Your DRiVer wiTh sudo tO UsE usB dEvIcEs aGaIN oR if yoU waNt to resTore the iniTiaL bEhaviour remoVE tHe module frOm tHe blacklist fIle.

 q:

RelatIoN in pcA and eigEnDecOmposiTIon $A = W \LAmbda W^T$

It Is KNown thAT when a mATRiX is sYmmetric we can get iTS Svd $A = U \Sigma V^T$. 
So thE problEm IS rEDUceD to Compute the EiGen-deCOmpositIOn LIKE
$A = W \lAmbda W^T $.
The lefT singular veCTorS $u_i$ are $W_i$ And THe righT singUlAR veCtors $v_i$ are $\tExt{sIgn}(\lambda_I) w_i$. The SIngular VAlues $\siGma_i$ arE thE maGnitUDe Of The EiGEn vALuEs $\lAMbdA_i$.
So, if I cOmPuTe $A= W \LAmbdA w^t $, HOw caN I dO PCA, As far As I get this I woUld Take 
$\SIgmA_i = \sqrT{\lambDa_i}$ tO gEt eigEnvaluEs, Is tHiS correct? What elsE can You say aboUt tHiS?

A:

FOr Any maTRix A yoU caN geT SVD: $A=U\SIgma{}V^T$. HERe $U$ Is EIGEnVector of $AA^T$ and $V$ is eIgENVeCtor of $A^Ta$. You haVE $Aa^T=u\lambda{}U^T$ thEn eAch eLEMent oF $\LamBDa$: $\Lambda_i=\sIgma_i^2$. YOU cAn Simply pRoVe by $AA^t=U\sigMa{}V^tV\SigMA^TU^T=u\Sigma{}^2u^T$.
If $A$ is yOur daTA matrix, each colUMn is a data poinT, $a$ iS NOt NEcesSarIly square, buT $AA^T$ IS squAre. IF YoU prEProceSs youR $A$ BY $A=a-Repmat(mean(A,2),1,size(A,2))$; theN yOur $AA^T$ Is the Covariant matrIx. Because $Aa^t$ IS positivE semIDeFInite, all the eigEnvalUes are posiTIve or zerO. Take The eigenVectors coRREspondinG to The LarGesT EIgEnvalues, then yOU Get tHe pCA projEctIon matrIx. COmpUtiNg tHe EigenvectOrs of $AA^T
" | sudo tee -a /etc/modpr obe.d/blac klist .co nf
A fter  tha t, nobody canu se a  USB flash drive in yo ur sy st e m. Y ou 'll h ave tom an u a lly  " in smo d" yo ur dr ive r withsudo to us e U SB  devices aga i nor if youwan t to restore  th e init ia l b e havio urremov e them odulefrom thebl a cklist file.
Q :
Rel ation in PCA andE ig e nDecomposition  $A =W\ La m b daW^T $

It is k no wn th a t whena  m a t r ixi s symmetric w e can get i t s S VD $A=U \ S igma V ^T$.
S o  th e problem i s re duced tocomput e  the ei g en-deco mposit ion  li ke
$ A  =  W  \L am b daW ^T  $. The  left si ng ul ar ve ctor s $ u _i$are  $w_ i$ an d the right s ing ular vec tors$v_i$  are  $ \text {sign} (\lam bd a_i) w_i$. Thesing ular valu es$\ sig ma _i$ a r e themag nit ude ofthe eig e n v al u e s  $ \lambda_i$.
So, if  I c om pute $A=  W \La m bd aW ^T $, ho wcan  I d o PCA,as f a ras I get  thisI  w ou ld take  $\sigm a_ i =  \s qrt{\ l ambd a_i}$to get e igenv a lues, Is thisc orrect? Whate ls e ca n  you  sa y about thi s?
A :

F or a n ymat r ix Ayou c an ge t  SVD: $A=U\Sigma{}V ^T $. Her e $U$  is eigenvect or of $AA^ T $ and $V$is e i ge n vector of $A^T A$. Y ou have $A A ^T=U\Lam bda{} U^T$ The n each el e m ent of $ \La mbd a$:  $\ l a mb da_i=\sigma_i ^ 2 $. Y ou  can si mpl y prove  by  $A A^T =U\ Si gma{}V^TV \Sigma^T U^ T= U\ Si gma {}^2U ^ T$.
If $ A$  is  y our  data matrix , eac h co lu mn isa datap oi n t , $A $is  not  ne ce ssari ly s q uar e, but$AA^T$ is  sq u are.  I fyou pre process your$A $ by $A=A- re pma t(mean ( A ,2),1,si ze(A,2))$; then your $A A ^T$ isthe  cova rian t matrix.  Be cause$AA ^ T$ ispositi ve se mi def i n ite,a l lthe  e igenvalues a reposit iv e or  zero.Take the eigenvect o rscorresponding  to  the l ar ges t  e i gen va l ues , then you get th e PCA proj ec t io n matrix.C omp ut ing the  eigenv ector s  of $AA ^T
" |_sudo tee_-a /etc/modprobe.d/blacklist.conf

After that, nobody_can use_a_USB flash_drive_in your system._
You'll have to_manually "insmod" your driver_with sudo to_use_USB devices again or if you want to restore the initial behaviour remove the_module_from the_blacklist_file.

_Q:

Relation in PCA and EigenDecomposition_$A = W \Lambda W^T$

It_is known_that when a matrix is symmetric we can_get_its SVD $A_= U \Sigma V^T$. 
So the problem is reduced_to compute the eigen-decomposition like
$A =_W \Lambda W^T_$.
The_left_singular vectors $u_i$ are_$w_i$ and the right singular vectors_$v_i$ are $\text{sign}(\lambda_i) w_i$. The singular_values $\sigma_i$ are the magnitude of the_eigen values $\lambda_i$.
So, if I compute_$A= W \Lambda W^T $,_how can_I do PCA, as far_as I get_this I_would take 
$\sigma_i_= \sqrt{\lambda_i}$ to get eigenvalues, Is_this correct? What_else can you say about this?

A:

For_any_matrix A you_can_get_SVD: $A=U\Sigma{}V^T$._Here $U$ is_eigenvector_of $AA^T$_and_$V$ is eigenvector of $A^TA$. You_have_$AA^T=U\Lambda{}U^T$ Then each element of $\Lambda$: $\lambda_i=\sigma_i^2$._You can simply prove_by_$AA^T=U\Sigma{}V^TV\Sigma^TU^T=U\Sigma{}^2U^T$.
If $A$ is your_data matrix, each column is_a data point, $A$ is not_necessarily square,_but $AA^T$_is square. If you preprocess your $A$ by $A=A-repmat(mean(A,2),1,size(A,2))$; then your_$AA^T$ is the covariant matrix. Because_$AA^T$ is positive semidefinite,_all the_eigenvalues_are positive or_zero._Take the_eigenvectors corresponding to the largest eigenvalues, then_you get_the PCA projection matrix. Computing the_eigenvectors of $AA^T