Here, matrices $P,Q \in \mathbb{R}^{n \times n} $ are satisfied with ${\rm rank}(Q)=r$, ${\rm rank}(P)\geq r$. I approximate $P$ as $Q$ in terms of Frobenius norm.
$$
Q' = \underset{Q \, : \,\operatorname{rank}(Q)=r}{\operatorname{arg min}} \| P – Q \|_F
$$
My question is below: is $Q'$ unique to given matrix $P$ ?
Supplement
I know Eckart-Young Theorem. According to this theorem,
$$
|| P – Q' ||_F = \sqrt{\sigma_{r+1}^2 + \cdots + \sigma_{n}^2 }.
$$
$\sigma_i$ is $i$-th biggest singular value of $P$. We can get $Q'$ by singular values decomposition of $P$. When we find $r$ rank matrix $Q$ which is satisifed with $|| P – Q ||_F = \sqrt{\sigma_{r+1}^2 + \cdots + \sigma_{n}^2 } $, is $Q$ same as the matrix which is given by $r$ rank approximation by singular values decomposition?
EDIT
Now, I understand $Q'$ is not unique to $P$.
When we find $r$ rank matrix $Q$ which is satisifed with $|| P – Q ||_F = \sqrt{\sigma_{r+1}^2 + \cdots + \sigma_{n}^2 } $, how can I find $Q$ is same matrix as $r$-rank approximation of $P$ by SVD or not?
Best Answer
Let $\pi$ be an orthogonal projection of rank $r$ that commutes with $P^{\mathrm t}\!P$ then $Q = P \pi$ is a critical point of the function $$A \mapsto \lVert P - A \rVert^2$$ over all $A$ of rank $r$. Moreover $$\lVert P - Q \rVert^2 = \operatorname{tr}(P^{\mathrm t}\! P\cdot (1-\pi))$$ and this is minimal when $\pi$ projects to the eigenspaces of $P^{\mathrm t}\! P$ with highest eigenvalues (so $1-\pi$ projects to the eigenspaces with lowest eigenvalues).
Uniqueness of $\pi$ depends on the eigenspace decomposition of $P^{\mathrm t}\! P$ and in particular the multiplicities of its eigenvalues. For example, if the largest $r$ eigenvalues (counting multiplicity) are all distinct then $\pi$ is unique. On the other extreme, when $P=1$ then all rank $r$ projections give the minimal distance.