How can a transformation be linear transformation without linearity? My teacher at the University gave me a question I could not understand completely. Here is the question:


  Let $T: \mathbb R^3 \to P[x]$ be a linear transformation with $$T([1, 0,
0])=x+1, \quad T([0, 1, 0])=x^2-x, \quad T([0, 0, 1])=x^2,$$ find
  $T([a, b, c])$, also find the standard matrix $A$ for the
  transformation.


The part that I did not understand is that how can $T([0, 1, 0])$ and $T([0, 0, 1])$ be linear since they have $x^2$. Also the $T([1, 0, 0])$ term has a constant. Those violate the linear transformation rules. Don't they?
As

$$[a,b,c]=a[1,0,0]+b[0,1,0]+c[0,0,1],$$

by linearity

$$T([a,b,c])=aT([1,0,0])+bT([0,1,0])+cT([0,0,1])=a(x+1)+b(x^2-x)+cx^2.$$

The linearity is not on $x$, but on $a,b,c$.
Linearity is nothing more or less than requiring  $T(u+v)=T(u) +(v), T(au)=aT(u)$. This does not rule out
$T(u)=x+1$. 

In fact consider the function $M\colon P[x]\to P[x]$ which has the effect of multiplying anything by $x^2+3x+1$ (or any random, fixed polynomial).

$M(f(x)+g(x) )= M(f(x) ) + M(g(x))$. ANd M(af(x) ) = aM(f(x)).

Can see  $M(1)= x^2+3x+1$, a quadratic function! $M$ satisfies the requirement of linear transformation, Perfectly alright.