Best orthogonal approximation of rank 1 matrix

Let $$X=\lambda_0u_0v_0^T\in\mathbb{R}^{n\times n}$$ be a rank 1 matrix where $$\lambda_0\in\mathbb{R}$$, $$u_0,v_0$$ are of unit Euclidean norm. What is the solution of the following problem? $$\hat{X}=\arg\min_{\substack{Y:Y=aU\\U\in O(n)\\a\in\mathbb{R}}}\|Y-X\|_F^2$$ where $$U$$ is orthogonal and $$\|\|_F$$ is the Frobenius norm. In other words, what is the best scaled orthogonal approximation of rank 1 matrix?

• Just an attempt at a solution... You can assume WLOG that $X$ has only one nonzero entry $X_{11}>0$ by taking its SVD; then I'd guess that the minimizer is $\frac{\|X\|}{n} I_n$, and I would suggest looking into Weyl's inequalities for singular values (Exercise 22 here) to prove it. – Federico Poloni Jun 10 at 22:18
• Isn't this an instance of the orthogonal Procrustes problem? – Rodrigo de Azevedo Jun 11 at 6:33
• @RodrigodeAzevedo No, because of the $a$ term. But I guess that one can solve the problem for every fixed $a$ using the orthogonal Procrustes result, and then take the minimum of the resulting single-variable function. – Federico Poloni Jun 11 at 6:48

As suggested by Federico Poloni in a comment, it suffices via the SVD to consider the case when $$X$$ has $$x_{1,1} > 0$$ and $$x_{i,j} = 0$$ when $$(i,j) \neq (1,1)$$. Then for any orthogonal matrix $$U$$ with columns $$\mathbf{u}_1,\ldots,\mathbf{u}_n$$ and scalar $$a \in \mathbb{R}$$ we have \begin{align*} \|X - aU\|_F^2 & = \|x_{1,1}\mathbf{e}_1 - a\mathbf{u}_1\|^2 + \sum_{j=2}^n\|a\mathbf{u}_j\|^2 \\ & = \big(x_{1,1}^2 - 2ax_{1,1}u_{1,1} + a^2\big) + \big(\sum_{j=2}^na^2\big) \\ & = x_{1,1}^2 - 2ax_{1,1}u_{1,1} + na^2, \end{align*} where $$\mathbf{e}_1 \in \mathbb{R}^n$$ is the first standard basis vector. There is only one term in this expression that depends on $$U$$ at all, so (for any given value of $$a$$) it is minimized when we make $$u_{1,1}$$ as large as possible (i.e., $$u_{1,1} = 1$$). The remaining columns of $$U$$ are arbitrary (but of course must be orthogonal to the first column $$\mathbf{e}_1$$).
Our goal is then to minimize \begin{align*} x_{1,1}^2 - 2ax_{1,1} + na^2, \end{align*} which is a quadratic in $$a$$ with vertex (i.e., minimum) at $$a = x_{1,1}/n$$.