Showing some transformation is linear Let $T: P_3(\mathbb{R}) \to P_3(\mathbb{R})$ be an operation defined by
  $$T(a+bx+cx^2+dx^3) = a + dx + (a+d)x^2  +(b-c)x^3$$
  Show that $T$ is linear


What I have done so far is look at it like this(since such transformations are homomorphic):

$$T(a+bx+cx^2+dx^3) = aT(1) + bT(x) + cT(x^2) + dT(x^3)$$

Now it would seem that the transformation does nothing, and rather the coefficients have been rearranged.

$\left( \begin{align} a\\b\\c\\d \end{align} \right)$ = $ a\left( \begin{align} 1\\-1\\0\\-1 \end{align} \right)$

Does this show that $T$ is linear? Or what am I meant to do?
It is not quite clear what you have done but then this is what you are supposed to do :

You have to show that $T(a+bx+cx^2+dx^3)=aT(1)+bT(x)+cT(x^2)+dT(x^3)$

It is given that $$T(a+bx+cx^2+dx^3) = a + dx + (a+d)x^2  +(b-c)x^3$$

So, we have :


$T(1)=T(1+0\cdot x+0 \cdot x^2+0\cdot x^3)=1+0\cdot x+(1+0)x^2+(0-0)x^3=1+x^2$


So, $T(1)=1+x^2$ 


$T(x)=T(0+1\cdot x+0 \cdot x^2+0\cdot x^3)=0+0\cdot x+(0+0)x^2+(1-0)x^3=x^3$


So,$T(x)=x^3$


$T(x^2)=T(0+0\cdot x+1 \cdot x^2+0\cdot x^3)=0+0\cdot x+(0+0)x^2+(0-1)x^3=-x^3$


So, $T(x^2)=-x^3$


$T(x^3)=T(0+0\cdot x+0 \cdot x^2+1\cdot x^3)=0+1\cdot x+(0+1)x^2+(0-0)x^3=x+x^2$


So, $T(x^3)=x+x^2$

Now, can you show that :

$$T(a+bx+cx^2+dx^3)=aT(1)+bT(x)+cT(x^2)+dT(x^3)$$

P.S : I have computed only $T(1)$ before and user142198 understood this and accepted this answer. But then another user Exam in 5 days was having issues in computing $T(x),T(x^2),T(x^3)$ and so i have computed that Only for him..
Given

$$
T\Big( a + bx + cx^2 + dx^3 \Big) = a + dx + ( a + d ) x^2 + ( b - c ) x^3
$$

You need to show that $T$ is linear, thus

$$
T\Big( a P_1 + b P_2 \Big) = a T\Big( P_1 \Big) + b T\Big( P_2 \Big),
$$

where

$$
P_k = a_k + b_k x + c_k x^2 + d_k x^3.
$$

and it is clear that $T$ is linear.



Note that

$$
T( 1 - x - x^3) = 1 - x - x^3
$$

and

$$
T( x + x^2 ) = 0
$$