If you find any mistakes, please make a comment! Thank you.

## Solution to Mathematics for Machine Learning Exercise 7.7

Consider the quadratic program illustrated in Figure 7.4, $$\min _{\boldsymbol{x} \in \mathbb{R}^{2}}\quad \frac{1}{2}\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]^{\top}\left[\begin{array}{ll} 2 & 1 \\ 1 & 4 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]+\left[\begin{array}{l} 5 \\ 3 \end{array}\right]^{\top}\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]$$ $$\text { subject to }\quad\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \\ 0 & 1 \\ 0 & -1 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right] \leqslant\left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \end{array}\right]$$Derive the dual quadratic program using Lagrange duality.

Solution: The process is already in Chapter 7.3.2. I will give the answer directly from (7.52) on page 242.

The dual quadratic program is $$\min _{\boldsymbol{\lambda} \in \mathbb{R}^{4}}\ -\frac{1}{14}\left(\left[\begin{array}{l} 5 \\ 3 \end{array}\right]+\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \\ 0 & 1 \\ 0 & -1 \end{array}\right]^\top\left[\begin{array}{c} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda_4 \end{array}\right]\right)^\top\left[\begin{array}{ll} 4 & -1 \\ -1 & 2 \end{array}\right] \left(\left[\begin{array}{l} 5 \\ 3 \end{array}\right]+\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \\ 0 & 1 \\ 0 & -1 \end{array}\right]^\top\left[\begin{array}{c} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda_4 \end{array}\right]\right)-\left[\begin{array}{c} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda_4 \end{array}\right]^\top \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \end{array}\right]$$ $$\text { subject to }\quad\left[\begin{array}{c} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda_4 \end{array}\right]\geqslant 0.$$