# Minimize Sum of a Quadratic Matrix Form

For given PSD symmetric Matrix $$Q$$ and two real $$a$$ and $$b$$ i want to minimize the non linear form defined as:

$$\mathop {\min }\limits_{{\alpha _i},{u_i}} \sum\limits_{i = 1}^{n = N} {{\alpha _i}u_i^TQ{u_i}}$$

st: $$\begin{array}{l} U = [{u_1},{u_2}...,{u_N}]\\ U{U^T} = I\\ 0 < a < {\alpha _i} < b \end{array}$$

• $u_i^\top Qu_i\geq 0$ for all $i$, so you want $\alpha_i = a$ for all $i$.
– Surb
Mar 8, 2020 at 20:38
• also, for $N=1$, $u_1$ is the eigenvector of $Q$ corresponding to the smallest eigenvalue
– Surb
Mar 8, 2020 at 20:41

## 1 Answer

According to @Surb comments, we need $$\alpha_i = a$$ for all $$i$$. Thus I reformulate the problem here: $$$$\begin{array}{cl} {\min_{U}} & {a \cdot \operatorname{tr} (U^TQU)} \\ {\mathrm{s.t.}} & {U^TU = I} \end{array}$$$$ where $$U = \begin{pmatrix} u_1 \\ \vdots \\ u_N\end{pmatrix}$$.

This is a non-convex problem. Applying the Lagrangian Multiplier Method, we have $$$$\max_{V} \min_{U} a \cdot \langle U, QU \rangle - \langle V, U^TU-I \rangle.$$$$ The optimal solution is $$$$a \cdot Q\bar{U} = \bar{U}\bar{V}, \quad \bar{U}^T\bar{U} = I.$$$$ Thus $$V$$ is a diagonal matrix formed from the eigenvalues of $$Q/a$$. And $$U$$ is the matrix formed from the eigenvector of $$Q$$.

In conclusion, if $$N < n$$, then $$V$$ is a diagonal matrix formed from the largest $$N$$ eigenvalues of $$Q/a$$, and $$U$$ is the matrix formed from its corresponding eigenvector.