# Why the negative reward function in linear quadratic regulation encourage to be at original state?

I was reading Stanford's CS 229 materials on Linear Quadratic Regulation ( Lecture note 13, youtube Lecture 18, around time 36 min). And it mentions that:

" the quadratic formulation of the reward is equivalent to saying that we want our state to be close to the origin.For example, if $U_t = I_n$(the identity matrix) and $W_t = I_d$, then $R_t = -\Vert(s_t)\Vert^2-\Vert(a_t)\Vert^2$, meaning that we want to take smooth actions (small norm of $a_t$) to go back to the origin (small norm of $s_t$)."

but why is that? Why the norm of the origin is smaller? $s_t$ for any time point are just a vector of n dimension which can take any value, right?

• The norm of $s_t$ is smaller the closer it is to the origin. – Yuval Filmus Feb 18 '18 at 7:35

Minimizing $s^TUs=s^Ts=\|s\|^2$ minimizes the distance from the origin, because the norm of a vector is its Euclidean distance (https://en.wikipedia.org/wiki/Euclidean_distance) from the origin:
$$\|s\|= \sqrt{s_1^2+s_2^2+\dots+s_n^2} = \sqrt{(s_1-0)^2+(s_2-0)^2+\dots+(s_n-0)^2}$$