[Math] How to find coordinate vector for an ordered none standart basis of polynomials

linear algebralinear-transformations

I know that it's trivial to know the coordinate vector with respect to the standart basis of polynomials.
The standart basis of $P_2[R]$ is:

$$\{1,t,t^2\}$$

A general vector spanned by the above set is of the form:

$$a+bt+ct^2$$

such that $a,b,c \in R.$

the coordinate vector with respect to the standart basis of $R^3$ is:

$$\left( \begin{array}{l}a\\b\\c\end{array} \right) = a\left( \begin{array}{l}1\\0\\0\end{array} \right) + b\left( \begin{array}{l}0\\1\\0\end{array} \right) + c\left( \begin{array}{l}0\\0\\1\end{array} \right)$$

But what if I have for example this ordered basis of $P_2[x]$ :

$$B=\{1+t,t^2,t\}$$

And I have a linear transorfmation
$$T:{P_2}[R] \to {M_{2×2}}[R]$$
and I need to find
$$[T]_B^E$$

This is the matrix that represent $T$ with respect to the basis $B$ and $E$ the standart basis for ${M_{2×2}}[R]$.

How can I find the general form of the coordinate vector with respect to the basis $B$ ?

I can only work with the coordinate verctor since I'm looking for the representing matrix for $T$ with respect to this basis.

so I need to find:

$$\begin{array}{l}T({b_1})\\T({b_2})\\T({b_3})\end{array}$$

And I dont know How to find them because Ineed the coordinate vector with respend to the basis $B$.

If it was for example to find $T[b_i] $ with respect to the standart basis of $P_2[R]$ ,
I would perform :

$$\begin{array}{l}T\left( \begin{array}{l}1\\0\\0\end{array} \right)\\T\left( \begin{array}{l}0\\1\\0\end{array} \right)\\T\left( \begin{array}{l}0\\0\\1\end{array} \right)\end{array}$$

But B is not the standart basis for $P_2[R]$ so I cant do $T$ on the three vectors above.

Best Answer

Suppose you have a polynomial $p + qt +rt^2$, and you want to find its coordinates with respect to the basis $\{1+t,t^2,t\}$ that you mentioned. In other words, you want to find numbers $a$, $b$, $c$ such that $$ p + qt +rt^2 = a(1+t) + b(t^2) + c(t) $$ Rearranging the right hand side , we get $$ p + qt +rt^2 = a + (a+c)t + bt^2 $$ For two polynomials to be equal, the coefficients of each power of $t$ must be equal, so \begin{align} p &= a \\ q &= a+c \\ r &= b \end{align} Now solve for $a$, $b$, $c$.