Determine the image of the polynomial $p(t)=(t-2)^2$ through the linear the transformation $T$

Consider $T:\mathbb{P}_2 \rightarrow \mathbb{P}_2 $ the linear transformation given by:

$T(t^2)=t^2+2t$

$T(t+1)=t-2$

$T(t)=1$

Determine the image of the polynomial $p(t)=(t-2)^2$ through the linear the transformation $T$, in the form of:

$T(p(t))=at^2+b+c$

I'm currently stuck on how do deal with the linear transformations of the polynomial, such as $T(a^2)=b$ or $T(a+1)=c$ and as such. Any help would be appreciated.

Thanks in advance.

You know that:

• $T(t^2)=t^2+2t$;
• $T(t)=1$;
• $T(1)=T\bigl((t+1)-t\bigr)=T(t+1)-T(t)=t-3$.

Therefore\begin{align}T\bigl((t-2)^2\bigr)&=T(t^2-4t+4)\\&=T(t^2)-4T(t)+4T(1)\\&=t^2+2t-4+4(t-3)\\&=t^2+6t-16.\end{align}

I see that Jose Carlos Santos answered your question elegantly. Hopefully, this answer adds a new way of seeing how to solve this problem with mapping matrices.

Let $B$ be the basis for $P_2$ formed by the vectors $b_1=t^2,b_2=t+1$ and $b_3=t$. $B=\{t^2,t+1,t\}=\{b_1,b_2,b_3\}$. Using the information given we know:

$T(b_1)=t^2+2t=1b_1+0b_2+2b_3$

$T(b_2)=t-2=0b_1-2b_2+3b_1$

$T(b_3)=1=0b_1+b_2-b_3$

With what we found previously,we can create a mapping matrix with respect to the B-basis (from B to B).

$_BF_B=\begin{pmatrix}1&0&0\\0&-2&1\\2&3&-1\end{pmatrix}$

All that is left is to find the b-coordinate vector for $(t-2)^2$ and multiply it with the mapping matrix we've created. $(t-2)^2=t^2-4t+4=b_1+4b_2-8b_3$

The b-coordinate vector is $\begin{bmatrix}1\\4\\-8\end{bmatrix}$

Computing the matrix vector product:

$\begin{pmatrix}1&0&0\\0&-2&1\\2&3&-1\end{pmatrix}\cdot\begin{pmatrix}1\\4\\-8\end{pmatrix}=\begin{pmatrix}1\\-16\\22\end{pmatrix}$

So $T((t-2)^2)=b_1-16b_2+22b_3$

$=t^2-16(t+1)+22t=t^2+6t-16$