Frobenius norm: completing squares and minimizing

Question

I would like to minimize the following quantity:

$Q = \left\lVert{X - C}\right\rVert^2_F + a\left\lVert{X - I}\right\rVert^2_F$

Where $X\in\mathbb R^{n\times n}$ is unknown, $C\in\mathbb R^{n\times n}$ is a known positive semi-definite and symmetric matrix, $I$ is the identity matrix, $a\in\mathbb R^+$ and $\left\lVert\cdot\right\rVert_F$ is the Frobenius norm. There is also some constraint on $X$, but for simplicity let's assume that it is only needed to be positive semi-definite. If I could somehow complete the squares on $Q$ then I could you use this answer to solve my problem.

Any help would be much appreciated. Thanks in advance.

Answer 1

The objective function is equal to $(a+1)\left\|X-\frac{C+aI}{a+1}\right\|_F^2+\text{constant}$. Hence the unique global minimiser is $X=\frac{C+aI}{a+1}$. As $C\succeq0$ and $a\ge0$, $X$ is positive semidefinite.

Answer 2

Addendum to user1551

Find the gradient and set it to zero, i.e., \begin{align} \frac{\partial Q}{\partial X} = 2(X - C) + a 2(X - I) = 0 \Longrightarrow X = \frac{C + a I}{a+1}. \end{align} The solution is exactly the same as user1551's solution.

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p.s.: there are several posts that explain how to compute gradient of Frobenius norm. For instance, say $f:= \|X\|_F^2 = \operatorname{tr}(X^T X)\equiv X:X$ (double colon to denote Frobenius product). Then, compute differential followed by the gradient. So, $df = dX:X + X:dX = 2X:dX \Longrightarrow \frac{\partial f}{\partial X} = 2X$.