Prove that linear functions f(x)=Ax are unique

Question

By definition, a function $f:\Bbb R^n \to \Bbb R^m$ is linear if

1. $f(x+y)=f(x)+f(y) \forall x,y\in \Bbb R^n$
2. $f(ax)=af(x) \forall x\in \Bbb R^n$

I want to prove that $f$ is linear iff $f(x)=Ax,A\in\Bbb R^{m\times n}$ and A is unique for any x.

I try to prove it by showing $f(x)=f(x\cdot1)=xf(1)=ax$ if $f$. But $x$ and $1$ is not in the same dimension. How can I do it?

And to prove that $A$ is unique, it means if

$Ax=Bx, 0=f(x)-f(x)=Ax-Bx=(A-B)x$, then $A=B$.

Am I understanding it right?

Answer 1

We need to prove two implications

1. if $f$ is linear $\implies$ $f(x)=Ax$
2. if $f(x)=Ax$ $\implies$ $f$ is linear

To prove "1" you need to show that every $x$ can be expressed by a linear combination of a basis $x=\sum a_i\cdot v_i$ and that $f(x)$ for linearity is completely determinated by $f(v_1), f(v_2),...,f(v_n)$.

The unicity of A, for a given basis can be shown assuming that for $v\neq w$ $$f(v)=f(w) \iff f(v)-f(w)=0\iff f(v-w)=0\iff v-w=0\iff v=w$$

To prove "2", assuming $$f(x)=Ax,A\in\Bbb R^{m\times n}$$

we need to show that

1. $A(x+y)=Ax+Ay \quad \forall x,y\in \Bbb R^n$
2. $A(ax)=aAx \quad\forall x\in \Bbb R^n$

which are true for matrix properties. Unicity also is trivial to prove.

Answer 2

To prove f is linear: Two parts in your definition are equivalent to say $f(ax+by)=af(x)+bf(y)$.

Now, $f(ax+by)=A(ax+by)=aAx+bAy=af(x)+bf(y)$. So, it is linear.

I think your proof on uniqueness is acceptable. Just make sure to check the case of $x=0$.