Find the coordinate vector of $p(t)=2+3t-2t^2$ in shifted basis $\{1,t-2,(t-2)^2\}$ How can I solve this? or can I have some references to look at?
$$p(t)=2+3t-2t^2$$ in shifted basis $$\{1,t-2,(t-2)^2\}$$
 A: Consider the polynomial as a vector object with coordinates $\langle 2,3,-2 \rangle$ in a vector space with a basis given by $ \{1,t,t^2\}$. Your goal is to to rewrite the polynomial in the form $$  a + b(t-2) + c(t-2)^2$$ and report the coefficients $a,b,c$ as a coordinate vector of the form $\langle a,b,c \rangle$- the coordinates with respect to the new basis. 
In addition to the linear algebraic methods for changing basis, you can rewrite the polynomial in the required form by a judicial application of completing the square and factoring, or by a very simple application of Taylor's theorem to discover the coefficients of the Taylor series centered at $t = 2$ (this is probably the most efficient method). Alternatively, you could expand the polynomial above and solve the system of equations for the undetermined coefficients. 
A: I see Andrew already has a good answer. You can also solve this as a matrix equation system if you expand the polynomials and turn them into column vectors and expanding $(t-2)^2 = t^2 - 4t + 4$ :
$${\bf T}= \left[\begin{array}{rrr}
1&-2&4\\
0&1&-4\\
0&0&1
\end{array}\right]$$
You see which numbers are to be placed where?
And then we solve for $\bf v$ the equation system: $${\bf T}{\bf v} =  \left[\begin{array}{r}2\\3\\-2\end{array}\right]$$
This will be easy once you learn how to solve matrix equation systems.
To expand on JnxF's solution (to help avoid any confusion) the matrix used to multiply in his answer is in fact the inverse of $\bf T$:
$${\bf T}^{-1} =  \left[\begin{array}{ccc}
1&2&4\\
0&1&4\\
0&0&1
\end{array}\right]$$
What is done when solving the above equation system: If we multiply to the left both sides with ${\bf T}^{-1}$:
$${\bf T}{\bf v} =  \left[\begin{array}{r}2\\3\\-2\end{array}\right] \Leftrightarrow \underset{= {\bf Iv} = {\bf v}}{\underbrace{{\bf T}^{-1}{\bf T v}}} = {\bf T}^{-1}\left[\begin{array}{r}2\\3\\-2\end{array}\right] $$
If you analyze ${\bf T}^{-1}$ and do completing of the square you will find that it corresponds to $\{1,t+2,(t+2)^2\}$ so it actually is "moving" the coordinate system back: $t\to t+2$ which undoes the previous change $t \to t-2$.
A: Another approach: basis conversion.


*

*Take $B = \{ 1,\; t,\; t^2\}$ the canonic base of $P_2(\mathbb R)$.

*And let $B' = \{ 1,\; t-2,\; (t-2)^2\} = \{ 1,\; t-2,\; t^2 -4t + 4\}  $
As $p(t) = 2 + 3t - 2t^2 \in P_2(\mathbb R)$, we can find its coordinates in the canonic base with no work:
$$p(t)_B = \left(\begin{matrix}2\\3\\-2\end{matrix} \right)$$
Now, apply change of basis and you're done.
It is not difficult to see that $$(1)_{B'} = \left(\begin{matrix}1\\0\\0\end{matrix} \right),\qquad (t)_{B'} = \left(\begin{matrix}2\\1\\0\end{matrix} \right), \qquad (t^2)_{B'} = \left(\begin{matrix}4\\4\\1\end{matrix} \right),$$ so:
\begin{align}p(t)_{B'} &= P^B_{B'} \cdot p(t)_B \\[1em]&= 
\left(\begin{matrix}\vdots&\vdots&\vdots \\ (1)_{B'} & (t)_{B'} & (t^2)_{B'} \\\vdots&\vdots&\vdots \end{matrix}\right) 
 \cdot p(t)_B \\[1em]&= \left(\begin{matrix}1&2&4\\0&1&4\\0&0&1\end{matrix}\right) \cdot \left(\begin{matrix}2\\3\\-2\end{matrix} \right) = \left(\begin{matrix}0\\-5\\2\end{matrix} \right)\end{align}
So you can check that 
$$ \color{red}{0} \cdot 1 + \color{red}{(-5)} \cdot (t-2) + \color{red}{(-2)} \cdot (t-2)^2 = p(t)$$
