Orthogonal projection to symmetric matrices with bounded eigenvalues

Let $$0, $$\mathbb{S}^n$$ be the set of all $$n\times n$$ real symmetric matrices, and $$\mathcal{V} \subset\mathbb{S}^n$$ be the set of symmetric matrices such that $$aI\le X \le b I$$, where $$\le$$ means the Loewner order. Consider the orthogonal projection $$\Pi:\mathbb{S}^n \to \mathcal{V}$$ such that for all $$A\in \mathbb{S}^n$$, $$\Pi(A)=\arg\min_{X\in \mathcal{V}}\|X-A\|^2.$$ where $$\|\cdot\|$$ indicates the Frobenius norm or spectral norm.

I was wondering whether there is a simple procedure to evaluate $$\Pi$$ (for one of the above matrix norms). Can the projection operator $$\Pi$$ be expressed in terms of projecting the corresponding eigenvalues of $$A$$?

As shown here, the claim holds if $$a=0$$ and $$b=\infty$$. The argument relies on any positive semidefinite matrices have nonnegative diagonals, which can not be easily extended to general $$a,b$$.

You can still reduce the problem to diagonals. For the Frobenius norm, von-Neumann's trace inequality gives

$$\|X - A\|_F^2 = \|X\|_F^2 + \|A\|_F^2 - 2 \langle X, A\rangle \geq \|X\|_F^2 + \|A\|_F^2 - 2 \sum_{i = 1}^n \sigma_i(X)\sigma_i(A) \\ = \| \Sigma_X \|_F^2 + \| \Sigma_A \|_F^2 - 2 \langle \Sigma_A, \Sigma_X \rangle = \| \Sigma_X - \Sigma_A \|_F^2$$

where $$\sigma_i(\cdot)$$ denotes the $$i$$-th singular value and $$\Sigma_A, \Sigma_X$$ are diagonal matrices holding the singular values of $$A$$ and $$X$$.

Note that the inequality is actually an equality when $$X$$ and $$A$$ commute; i.e., $$X$$ and $$A$$ have the same set of eigenvectors. Therefore, the optimal solution will be

$$X = U \mathbf{diag}\left(\mathrm{proj}_{[a, b]} \sigma_i(A)\right) U^*,$$

where $$U$$ is the matrix of eigenvectors of $$A$$.

• Thanks for your answer. May you kindly provide a reference for the equality condition of von-Neumann's trace inequality?
– John
Aug 5 at 23:18
• I don't have an easily accessible reference at hand, but one way to verify that the condition is sufficient for equality is to simply set $X$ equal to $U D U^*$ for any diagonal matrix $D$ and $U$ being the eigenvectors of $A$ and notice that it is equal to the lower bound. Aug 6 at 17:54
• To verify the claim, have you assumed that $A=U\Sigma_A U^*$? As $\Sigma_A$ contains the singular values of $A$, and $A$ is not necessarily positive semidefinite, it seems that $U$ is not the eigenvectors of $A$. In fact, I feel the correct formula is $\Pi(A)= U \mathbf{diag}\left(\mathrm{proj}_{[a, b]} \sigma_i(A)\right) V^*$, where $A=U\Sigma_A V^*$ is the singular value decomposition of $A$ (here a related post). Please kindly let me know if I overlooked anything.
– John
Aug 7 at 23:27
• May you kindly clarify whether you have assumed that $A=U \Sigma_A U^*$ for the SVD? I don't think this expression is true for general symmetric $A$, as it implies that $A$ is positive semidefinite. In particular, your expression of $X$ does not recover the solution in here if $A$ is negative definite.
– John
Aug 11 at 20:57
• Sorry, I missed your previous comment! Yes, I did overlook that possibility. However, in that case you can just replace the von-Neumann inequality with Fan's trace inequality, which is expressed in terms of eigenvalues instead of singular values. I can clarify in my original answer if you need more context. Aug 11 at 21:19