TITLE: Optimal symmetric rank-1 approximation
QUESTION [5 upvotes]: I want to find $\mathbf{x}$ that minimizes $\|A-\mathbf{x}\mathbf{x}'\|^2$ where $\|\cdot\|$ is Frobenius norm.
Differentiating with respect to $\mathbf{x}$ and setting to $\mathbf{0}$, I get 
$$\mathbf{x}\mathbf{x}'\mathbf{x}=A \mathbf{x}$$
Any idea how to proceed further?
(Per request, here's how I did the derivative)
Let $J=\|A-\mathbf{x}\mathbf{x}'\|^2=\text{tr}(Y'Y)$ where $Y=A-\mathbf{x}\mathbf{x}'$
To compute derivative use technique of differentials as in Magnus,Neudecker and here
$$dY(\mathbf{x})=-2\mathbf{x} \mathbf{dx}'$$
$$dJ(Y)=\text{tr}(2Y'dY)$$
substitute definition of $dY$ and $Y$ into above, simplify to get
$$dJ(\mathbf{x})=-4 \text{tr}((A-\mathbf{x}\mathbf{x}')'\mathbf{x}\mathbf{dx}')=4\mathbf{x}'(\mathbf{x}\mathbf{x}'-A)'\mathbf{dx}$$
The last expression is a canonical form from which the derivative can be identified as the expression before $\mathbf{dx}$, doing transpose and setting to $\mathbf{0}$ gives the equation above.

REPLY [3 votes]: I would like to show how the norm may be split as the sum of symmetric (Hermitian) and skew symmetric (skew Hermitian). 
Let $A = H + S$ with $H = H^*$ and $S = -S^*$. Then
\begin{align}
 \left(H+S\right)\left(H+S\right)^* &= HH^* + HS^* + SH^* + SS^* \\
 &=HH^* -HS + SH + SS^*\\
\end{align}
From the cyclic property of trace, $\operatorname{Trace}(HS) = \operatorname{Trace}(SH)$, and the middle terms cancel in the norm.
Thus we have
$$\Vert A \Vert^2 = \Vert H+S \Vert^2 = \Vert H \Vert^2 + \Vert S \Vert^2$$
The addition of $\mathbf{x}\mathbf{x}'$ then affects only the symmetric portion since it is symmetric. Therefore, the original problem may be simplified using full symmetry, even if $A$ is not itself symmetric, as Dineshdileep noted.