MHB What Are Linear Transformations and How Do They Work?

Petrus
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Hello,
this is something basic I have hard to understand and would like to have help!:)
this is a exemple from My book and I Dont understand the input!
"Let $$T: P_2->P_2$$ be the linear transformation defines by $$T(P(x))=p(2x-1)$$
I Dont understand how this work
$$T(1)=1, T(x)=2x-1, T(x^2)=(2x-1)^2$$
Edit: if i think correct they think like this $$T(1)=T(x^0)=(2x-1)^0=1$$
Regards,
$$|\pi\rangle$$
 
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Petrus said:
Hello,
this is something basic I have hard to understand and would like to have help!:)
this is a exemple from My book and I Dont understand the input!
"Let $$T: P_2->P_2$$ be the linear transformation defines by $$T(P(x))=p(2x-1)$$
I Dont understand how this work
$$T(1)=1, T(x)=2x-1, T(x^2)=(2x-1)^2$$

Regards,
$$|\pi\rangle$$

Hi Petrus!

I suspect that should be $$T(P(x))=P(2x-1)$$.

If so, then for your first example, you would have P(x)=1.
So T(P(x)) = T(1).
And P(2x-1) = 1.
Therefore T(1) = 1.

In the second example, you would have P(x)=x.
So T(P(x)) = T(x).
And P(2x-1)=2x-1.
Therefore T(x) = 2x-1.

In the third example. you would have $P(x)=x^2$.
So $T(P(x)) = T(x^2)$.
And $P(2x-1) = (2x-1)^2$.
Therefore $T(x^2) = (2x-1)^2$.
 
I like Serena said:
Hi Petrus!

I suspect that should be $$T(P(x))=P(2x-1)$$.

If so, then for your first example, you would have P(x)=1.
So T(P(x)) = T(1).
And P(2x-1) = 1.
Therefore T(1) = 1.

In the second example, you would have P(x)=x.
So T(P(x)) = T(x).
And P(2x-1)=2x-1.
Therefore T(x) = 2x-1.

In the third example. you would have $P(x)=x^2$.
So $T(P(x)) = T(x^2)$.
And $P(2x-1) = (2x-1)^2$.
Therefore $T(x^2) = (2x-1)^2$.
Thank you! Evrything is clear now! Have a nice day!:)
By the way your suspect is correct!;)
Regards,
$$|\pi\rangle$$
 
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