Why is (specifically) linear algebra so important in physics?

As someone who got introduced to linear algebra this past year, I noticed that we're working with a generalization of the function $y(x)=kx$ for larger dimensions. But the theorems I've been taught don't even include the possibility of what would be $y(x)=kx +n$ (notice the $n$) in $\mathbb{R} \rightarrow \mathbb{R}$ functions. Why is such a seemingly limited tool so important for physics?

(Please, correct me if I said something which is not accurate. Thanks!)

• maybe becaus ethe linear relation is often between cause and effect - and small / zero cause have small / zero effect. Or it may be a matter of choosing the right units. With Kelvin scale, pressure of a gas is proportional to temperature, with Celsius or Fahrenheit, there is an offset. – Hagen von Eitzen Apr 29 '18 at 21:02
• Imagine you have 20 $y$s and 320 $x$s. What would you do? The second case is called an affine map $y=Ax+b$ with $A$ being a matrix and the rest are vectors – percusse Apr 29 '18 at 21:02
• Divide an conquer. Non-linear problems are too complicated. So, you approximate them with a linear one (have you seen Taylor?). A further reduction, making a change of coordinates $y=k(x-n/k)$ turns into $y=kx'$ by putting $x'=x-n/k$. So, studying linear algebra you are studying the first problem that is both tractable, and rich enough to deserve being studied. – user553213 Apr 29 '18 at 21:08
• In QM, the state of current reality in represented by some vector in a some Hilbert space. Every physical observable is simply a linear operator on same space. – achille hui Apr 29 '18 at 21:14