Define $T: Poly_2 \ to\ Poly_2$ by $$T(at^2+bt+c)=3ct^2 +2at-b$$
1) Show that T is a linear transformation and give a matrix A for T with respect to the basis $B=\{t^2,t,1\}$.
2) Give a matrix $Q$ for $T$ with respect to the basis $B_1=\{t^2+1 , t-2, t+3\}$ directly
3) Lastly show that $PAP^{-1} = Q$ where $P$ is the coordinate change matrix.
For Part 1: Since $T(3t^2+2at−b) = T(3t^2)+T(2t)−T(b)$ and $T(r(3t^2+2at−b)= rT(3t^2+2at−b)$ because $rT(3t^2+2at−b) = T(r(3t^2))+T(r(2t))−T(rb)$. it's a linear transformation.But is there more to show? I feel like this is just the definition in a sense.
For part 2 Since it's a linear transformation, the vectors are linearly independent, therefor there exists an a,b,c for which we can get the zero vector. Meaning we have $a(t^2+1)+b(t-2)+c(t+3)=0$....please let me know if this is correct so far, because I'm stuck and would like to know if Im even going the right direction. Not sure where to bring $3t^2+2at-b$ in but I feel as if it should be in the factored form $(t+1) (3 t^2-1)$
A linear transformation $f: V \to W$ has two properties:
These can be tested at the same time by confirming that $f(\alpha x + y) = \alpha f(x) + f(y)$ for all $x,y \in V$ and $\alpha$ in your base field (probably $\Bbb R$ in your class).
So let's test your linear transformation: $$\begin{align}T\left[\alpha(a_1t^2+b_1t+c_1) + (a_2t^2+b_2t+c_2)\right] &= T\left[(\alpha a_1+a_2)t^2+(\alpha b_1+b_2)t+(\alpha c_1+c_2)\right] \\ &= 3(\alpha c_1+c_2)t^2 + 2(\alpha a_1+a_2)t-(\alpha b_1+b_2) \\ &= \alpha(3 c_1)t^2 + \alpha(2a_1)t -\alpha(b_1) + (3c_2t^2 +2a_2t-b_2) \\ &= \alpha(3 c_1t^2 + 2a_1t -b_1) + (3c_2t^2 +2a_2t-b_2) \\ &= \alpha T(a_1t^2+b_1t+c_1) + T(a_2t^2+b_2t+c_2)\end{align}$$
Thus $T$ is linear.
So first we need to determine the coordinates of the input vector $\mathbf x = at^2 + bt+c$ and the ouput vector $\mathbf y = 3ct^2+2at-b$. Hopefully you can see that $$[\mathbf x]_B = \pmatrix{a \\ b \\ c},\quad [\mathbf y]_B = \pmatrix{3c \\ 2a \\ -b}$$
So we just need the matrix $A$ which takes $\mathbf x$ to $\mathbf y$. Thus we figure out $$\pmatrix{? & ? & ? \\ ? & ? & ? \\ ? & ? & ?}\pmatrix{a \\ b \\ c} = \pmatrix{3c \\ 2a \\ -b}$$
Clearly the matrix that does the job is $$A = \pmatrix{0 & 0 & 3 \\ 2 & 0 & 0 \\ 0 & -1 & 0}$$
If you need I can walk you through it, but hopefully you're far enough in your linear algebra studies that you can figure out how that matrix was obtained.
This part is really tedious. What we need is the coordinates of the $T(t^2+1)$, $T(t-2)$, and $T(t+3)$ wrt to the basis $B_1$. First let's figure out how to represent an arbitrary vector in this basis. That's equivalent to figuring out $[\mathbf x]_{B_1}$. So let's do that.
$$\begin{align}\mathbf x &= at^2+bt+c = a'(t^2+1)+b'(t-2)+c'(t+3) \\ &= a't^2+a'+b't-2b'+c't+3c' \\ &= a't^2 +(b'+c')t+(a'-2b'+3c')\end{align} \\ \implies \begin{cases} a=a' \\ b=b'+c' \\ c=a'-2b'+3c'\end{cases} \\ \implies \pmatrix{a \\ b \\ c} = \pmatrix{1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & -2 & 3}\pmatrix{a' \\ b' \\ c'} \\ \implies \pmatrix{a' \\ b' \\ c'} = \frac 15\pmatrix{5 & 0 & 0 \\ 1 & 3 & -1 \\ -1 & 2 & 1}\pmatrix{a \\ b \\ c}$$
That last step required finding the inverse matrix of $$\pmatrix{1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & -2 & 3}$$
Thus we can see that $[\mathbf x]_{B_1} = \frac 15\pmatrix{5a \\ a+3b-c \\ -a + 2b + c}$
Now that we've found the coordinates of an arbitrary vector wrt $B_1$, let's figure out where the basis vectors of $B_1$ get mapped to:
$$\begin{align}T(t^2+1) &= 3t^2+2t \\ &= \frac 15(15)(t^2+1)+\frac 15(3+6-0)(t-2) + \frac 15(-3+4+0)(t+3) \\ &= 3(t^2+1) + \frac 95(t-2) +\frac 15(t+3) \\ T(t-2) &= -6t^2-1 = -6(t^2+1) -(t-2)+(t+3) \\ T(t+3) &=9t^2-1= 9(t^2+1)+2(t-2)-2(t+3)\end{align}$$
Thus $Q$ will just be the matrix that maps $\pmatrix{1 \\ 0 \\ 0}$ to $\pmatrix{3 \\ \frac 95 \\ \frac 15}$, etc.
Then $$Q = \pmatrix{3 & -6 & 9 \\ \frac 95 & -1 & 2 \\ \frac 15 & 1 & -2}$$
What we want here is a matrix that'll change the coordinates of a vector with respect to the basis $B_1$ to the coordinates of that vector with respect to the basis $B$. Then once we've done that, the reason why $PAP^{-1}=Q$ is that on the LHA we're just converting the coordinates of an input vector from the $B_1$ basis to the $B$ basis, then using the matrix $A$ to perform the transformation $T$, then converting back to the $B_1$ basis. That is equivalent to just finding a single matrix which performs the transformation $T$ with respect to the basis $B_1$ -- i.e. the RHS should just be the matrix $Q$ found in part $2$.
All we need to do is figure out the coordinates of the basis vectors in $B_1$ with respect to $B$. That should be easy enough:
$$\begin{cases} t^2+1 = (1)t^2 + (1)1 \\ t-2 = (1)t+(-2)1 \\ t+3 = (1)t+(3)1\end{cases}$$
Therefore $$[t^2+1]_B = \pmatrix{1 \\ 0 \\ 1},\quad [t-2]_B = \pmatrix{0 \\ 1 \\ -2},\quad [t+3]_B = \pmatrix{0 \\ 1 \\ 3}$$
I posit that if we were to put those vectors into the columns of a matrix, then that would be the change of basis matrix from $B_1$ to $B$. Let's test it. Remember that $[t^2+1]_{B_1} = \pmatrix{1 \\ 0 \\ 0}$, so let's see where that maps to under this matrix:
$$\pmatrix{1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & -2 & 3} \pmatrix{1 \\ 0 \\ 0} = \pmatrix{1 \\ 0 \\ 1}$$ But that vector is just the coordinates of $t^2+1$ in the basis $B$. You can likewise confirm that this matrix converts the other two basis vectors in $B_1$ to their coordinates in $B$.
Thus $P^{-1}$ (in your question's notation) is $\pmatrix{1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & -2 & 3}$. Find its inverse to find $P$ and then take the product to see if you get back the matrix we found the hard way in part $2$.