linear systems - Does scaling property imply superposition?

For a system to be linear,it follow the principles of scaling and superposition.Does scaling imply superposition?If so why are two different conditions given for linearity?If not can u specify an example for which only one of them is satisfied.

Answer

The answer is "no" in general. Let us see the details.

The notion of linearity goes beyond systems. Let us start from vector, or linear spaces. They deal with two objects: vectors and "scalars". In a vector space, one can add vectors, and multiply a vector by a scalar. More formally, a vector space over a field $F$ is a set $V$ together with two operations, vector addition and scalar multiplication. Those operations in $V$ satisfy a number of axioms. The field $F$ possesses two operations as well, satisfying different axioms, namely field axioms, ruling the addition or product of scalars. Those two operations are often assimilated to addition and multiplication. But in general the addition/multiplication in $F$ are not the same as the vector addition/vector scaling in $V$. Indeed, if $f \in F$, $v\in V$, then $f.v \in V$, but is not anymore an element of $F$. The scalar has been cast to a vector. This plays a role in the standard confusion in defining linearity.

Now take for instance the field of rationals $F=\mathbb{Q}$ with the two standard operations. Take $V=\mathbb{R}$ the vector space of reals over $F$ with the standard real addition and real product. To relate this to the introduction, for instance, $1.\sqrt{2}$ is not rational anymore, although the multiplication seems to be the same, because you mix a scalar ($1$) and a vector $(\sqrt{2})$.

Suppose you have a system $S$ that provides the following output: if $v_1\in V$ is rational, $S(v_1) = v_1$; if $v_2\in V$ is irrational, $S(v_2) = -v_2$. For any rational $q$, $q.v_1$ is rational, and $q.v_2$ is irrational. Resultingly, $S(q.v_1) = q.v_1 = q.S(v_1)$, and $S(q.v_2) = -q.v_2 = q.S(v_2)$. Hence, $s(q.v) = q.S(v)$ for any $v\in V$, so $S$ verifies the scaling property.

However, $v_1+v_2 $ is irrational. Hence, $S(v_1+v_2) = -v_1 -v_2$, which is different in general from $S(v_1)+S(v_2) = v_1 - v_2$, take $v_1 = 1$ and $v_2 = \sqrt{2}$ for instance.

So scaling does not imply superposition (in your sense) in general.

But there exists somehow converse statements. In other domains, one sometimes calls the scaling "homogeneity", and with additivity we get the superposition principle for a system $S$: $S(q_1.v_1+q_2.v_2) = S(q_1.v_1)+S(q_2.v_2)$. Keeping with standard fields, it can be shown that additivity implies homogeneity with rational scalars, by playing on $S(v+v) = S(v)+S(v) = 2.S(v)$ etc. To go beyond rational scalar often requires additional assumptions like continuity, which could be troublesome depending on the sets you choose, see for instance Why doesn't superposition imply linearity? Why is homogeneity needed?