# orthogonal projection in projected gradient descent.

So I need to use gradient projection with the point $$x = (2,0)$$ and a step size of $$\frac{1}{2}$$ for the problem. \begin{equation} \begin{aligned} \min \quad & \frac{1}{2}x_1^2 + \frac{1}{2}x_2^2 \\ \textrm{s.t.} \quad & 2 - x_1 - x_2 = 0 \\ \end{aligned} \end{equation}

Projected gradient descent is defined as $$x_{k+1} = \prod_X (x_k - \tau_k \nabla f(x_k))$$ where $$\prod_X(x)$$ is orthogonal projection of $$x$$ on $$X$$ and $$\tau_k$$ is the step size.

So I attempted my first iteration and with $$x_k = (2, 0)^T$$ as my starting point $$\tau_k = \frac{1}{2}$$ as my step size. But this is the step where I am stuck $$\prod_X\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$. I don't know how to compute the orthogonal projection.

• What is the space $X$? $\prod_X(y)=\arg\inf_{x\in X}||y-x||$ – mm-crj Dec 14 '18 at 23:05
• Space $X$ is the constraint space and I found the orthogonal projection of $x$ on $X$, that is, the nearest point in $X$ to $x$. So if I plot $2 - x_1 - x_2 = 0$ as a line then the nearest point in $2 - x_1 - x_2 = 0$ to $(1 -1)^T$ is $(2,0)$ again which now my gradient doesn't go anywhere. – HD5450 Dec 14 '18 at 23:11