# Finding matrix associated with linear transformation

I need to find the matrix associated with the linear transformation :

$$T(a_{2}t²+a_{1}t+a_{0})=4(a_{2}-a_{0})t+2a_{1}$$

with respect to the basis B'(in the domain) and B(in the target)

where $$B=(1, 2t, 4t²-1)$$ and $$B'=(1,t,t²)$$

So the first thing I write is the required matrix, which is

$$M=[[T(1)B'][T(t)B'][T(t²)]]$$

The answer sheet states that I need to find

$$T(1)=-4t$$

$$T(t)=2$$

$$T(t²)=4t.$$

and then solve them in order to find the matrix columns and thus the required matrix

I know how to solve these three T's in order to find a matrix, however my doubt is:

Where are these numbers $$(-4,2,4)$$ coming from ? Why does it state that I need to find $$T(1)=-4$$ or $$T(t)=2$$?
I do not understand, for example, why it is equating $$T1=-4$$. Where is the minus $$4$$ coming from? Where is the $$2$$ from $$T(t)$$ coming from . This is my doubt, so an answer on why $$T1$$ the value of $$-4$$ or why $$T(t)$$ is equal to $$2$$ would be greatly appreciated.

I will add a link so that it may appear clearer. I am new here and any help on formatting would be great .enter image description here

• Read the answer sheet carefully; it says $T(1)=-4\color{red}{t}$, not $T(1)=-4$. Can you calculate $T(1)$ with the given formula of the linear transformation $T$? Nov 30 '18 at 15:12
• I'm sorry i mistyped. I do not understand what process to use to get -4t, can you please explain ?
– BM97
Nov 30 '18 at 15:14
• I have a form of dyscalculia and I am not able to connect the result -4 with T(1)
– BM97
Nov 30 '18 at 15:17
• Can you please show the calculations that result in -4t?
– BM97
Nov 30 '18 at 15:23

The given linear transformation maps a quadratic polynomial of the form $$a_2t^2+a_1t+a_0$$ to the linear polynomial $$4(a_2-a_0)t+2a_1$$, which is given to you by way of the formula: $$T\left(\color{blue}{a_2}t^2+\color{purple}{a_1}t+\color{green}{a_0}\right)=4(\color{blue}{a_2}-\color{green}{a_0})t+2\color{purple}{a_1}$$ If you want to find the image of, for example, the polynomial $$\color{blue}{2}t^2\color{purple}{-3}t+\color{green}{5}$$, note that this is exactly of the form $$\color{blue}{a_2}t^2+\color{purple}{a_1}t+\color{green}{a_0}$$ with $$\color{blue}{a_2=2}$$, $$\color{purple}{a_1=-3}$$ and $$\color{green}{a_0=5}$$ and hence the image becomes: $$T\left(\color{blue}{2}t^2\color{purple}{-3}t+\color{green}{5}\right)=4(\color{blue}{2}-\color{green}{5})t+2(\color{purple}{-3}) = -12t-6$$ Carefully study this example; I used coloring to help see what numbers go where.
If you got this, can you find $$T(1)$$? Notice that "$$1$$" is also a polynomial of the form $$a_2t^2+a_1t+a_0$$, namely with $$a_2=0$$, $$a_1=0$$ and $$a_0=1$$; and $$T$$ maps it to...?