# 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?

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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 $AA^T$ is the same as computing the SVD of $A$, so you can just compute the SVD of $A=U\Sigma{}V^T$, and take the first p columns of U as your projection matrix. Project your data $A$ to $U_{(:,1:p)}^TA$.
if i use correlation matrix of my data, let's say I convert a huge matrix to a $d \times d$ correlation matrix $\rho$, where $d$ is the dimension of the original matrix. If I do SVD of $\rho$ it would be equivalent to $\rho= \mathbf X \mathbf X^{\top}=\mathbf U \mathbf \Sigma^{2} \mathbf U^{\top}$ IS THAT CORRECT? – cMinor Nov 27 '12 at 6:11
Yes, because your $\rho$ is symmetric, the SVD is also symmetric. – Min Lin Nov 27 '12 at 6:16
one last question... What would be the relation of covariant matrix (you described) and a $d \times d$ correlation matrix when applying SVD to both of them? – cMinor Nov 27 '12 at 6:18