Largest eigenvector as the rank-1 approximation Consider a square real symmetric matrix $A$. The singular value decomposition (SVD) is given as
$A = U\Sigma U^T$,
and it can also be found by minimizing $|A-U\Sigma U^T|$ where $|.|$ is the $l$2 norm, $U$ is constrained to be orthonormal and $\Sigma$ is diagonal, or simply $A=UU^T$ if $U$ is orthogonal.
Now since $A$ is square, real and symmetric, is the SVD the same as the eigenvalue decomposition? I think so.
Now suppose I want to find the rank=1 approximation to A. I can take the first row of $U$ (corresponding to the largest element of $\Sigma$). This is the eigenvector with the largest (absolute) eigenvalue.
Is this the same as minimizing $| A - U_1 U_1^{T} |$ where $U_1$ is a $N$x$1$ row vector?
So is it true that the first eignevector $U_1$ is the solution to 
$\arg\min_{U_1} |A-U_1 U_1^{T}|$
Can I simply state this result as a known result in my work?
 A: In general, the SVD decomposition of a matrix $A$ is $A = U \Sigma V^T$ where $\Sigma$ is diagonal with non-negative entries (the singular values), the columns of $U$ are the eigenvectors of $A^T A$ and the columns of $V$ are the eigenvectors of $A A^T$. If $A$ is symmetric and positive semidefinite (so all the eigenvalues of $A$ are non-negative), then the SVD decomposition of $A$ coincide with the eigenvalue decomposition. If $A$ is symmetric but not necessarily positive semidefinite, the eigenvalue decomposition of $A$ is not the same as the SVD decomposition but quite close:
If you take a basis $(u_1, \ldots ,u_n)$ of eigenvectors of $A$ with $Av_i = \lambda_i v_i$ such that $|\lambda_1| \geq \ldots \geq |\lambda_n|$ then the SVD decomposition of $A$ can be constructed by choosing $U = (u_1 | \ldots | u_n)$ and $V = (\pm u_1 | \ldots | \pm u_n)$ (where you choose $-u_i$ if $\lambda_i < 0$ and $+u_i$ otherwise).
Finally, the matrix $\lambda_1 u_i u_i^T$ is indeed the best rank one approximation of $A$ in the sense that
$$ \lambda_1 u_1 u_1^T = \arg \min_{u,v} |A - uv^T| $$
where $| \cdot |$ is the Frobenius norm. You can read about this (and see the proof) here. It is well-known and you can find many sources if you need to refer to the result.
