How can a transformation be linear transformation without linearity?

Question

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?

Answer 1

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 $a,b,c$.

Answer 2

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.

Answer 3

How can $T([1,0,0])$ be linear? It's not linear; nobody said it was linear. Here $T([1,0,0])$ is a vector, and it makes no sense to say a vector is linear.

$T$ is linear, not $T(v)$ for a vector $v$.