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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

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  • $\begingroup$ 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 $T$? $\endgroup$
    – StackTD
    Nov 30 '18 at 15:12
  • $\begingroup$ I'm sorry i mistyped. I do not understand what process to use to get -4t, can you please explain ? $\endgroup$
    – BM97
    Nov 30 '18 at 15:14
  • $\begingroup$ I have a form of dyscalculia and I am not able to connect the result -4 with T(1) $\endgroup$
    – BM97
    Nov 30 '18 at 15:17
  • $\begingroup$ Can you please show the calculations that result in -4t? $\endgroup$
    – BM97
    Nov 30 '18 at 15:23
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I'm sorry i mistyped. I do not understand what process to use to get -4t, can you please explain ?

It looks like you don't know yet how to read such a formula for a linear transformation.

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...?

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  • $\begingroup$ THANKYOU!!!! I got it perfectly . $\endgroup$
    – BM97
    Nov 30 '18 at 15:31
  • $\begingroup$ Alright, you're welcome. $\endgroup$
    – StackTD
    Nov 30 '18 at 15:39

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