$$\begin{array}{ll} \text{minimize} & f(X)= \| A - XX^T \|_F^2 + \| X \|_F^2\\ \text{subject to} & X \in Gr(d,N) \end{array}$$
where $Gr(d,N)$ means the Grassmann Manifold; $\|\cdot\|_F$ is the Frobenius norm; $X \in R^{N \times d}$ and $A \in R^{N \times N}$ is symmetric.
My question is how to optimize above objective on Grassmann manifold? what step do we need?
Thanks.
We assume $d\leq N$ and $A$ symmetric $\ge 0$. If you use the construction of 3.1 in
http://openaccess.thecvf.com/content_cvpr_2017/papers/Wang_Grassmannian_Manifold_Optimization_CVPR_2017_paper.pdf
then your problem is equivalent to
$\text{minimize} \; f(X)= \| A - XX^T \|_F^2 + \| X \|_F^2\; \text{subject to} \; X^TX=I_d$.
Since you use the Frobenius norm, we can do explicitly the job. Indeed $f(X)=tr(A^2+2XX^T)-2tr(X^TAX)$; since $tr(A^2+2XX^T)$ is constant, the problem reduces to find
$\text{maximize} \; g(X)= tr(X^TAX)\; \text{subject to} \; X^TX=I_d$.
Let $spectrum(A)=\lambda_1\geq \cdots \geq \lambda_N\geq 0$.
$\textbf{Proposition}$. The max of $g$ is reached for $X=[u_1,\cdots,u_d]$ where $u_j$ is an eigenvector of $A$ associated to $\lambda_j$.
$\textbf{Proof}$. cf my post (EDIT 4BIS) in
Minimize ${tr(\mathbf{X}^T\mathbf{P}\mathbf{X})}/{tr(\mathbf{X}^T\mathbf{Q}\mathbf{X})}$ s.t. $\mathbf{X}^T\mathbf{X}=\mathbf{I}$
It suffices to adapt the proof of Proposition 2 when the order of the $(\lambda_i)$ is reversed.