# How to find coordinate vector for an ordered none standart basis of polynomials?

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 $$T:{P_2}[R] \to {M_{2x2}}[R]$$ 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 ${M_{2x2}}[R]$.

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.

• In this case it's easy, because the first element in the basis is the only one with a constant term, and the second element is the only one with a second-degree term. So you use as much of those as is necessary, and then you use the third to correct the resulting first-degree term. In general, however, it comes down to solving a system of three equations. Jan 24, 2016 at 12:58

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 $a$, $b$, $c$.

We are given the basis $B=\{b_1,b_2,b_3\}$ for $P_2(\Bbb R)$ where \begin{align*} b_1(t) &= 1+t & b_2(t) &= t^2 & b_3(t) &= t \end{align*} We are also given the basis $E=\{e_1,e_2,e_3,e_4\}$ of $M_{2\times 2}(\Bbb R)$ where \begin{align*} e_1 &= \left[\begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right] & e_2 &= \left[\begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array}\right] & e_3 &= \left[\begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array}\right] & e_4 &= \left[\begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array}\right] \end{align*} Finally, we are told that $T:P_2(\Bbb R)\to M_{2\times 2}(\Bbb R)$ is a linear map and we are asked to compute $[T]_B^E$.

To do so, note that $[T]_B^E$ is $$[T]_B^E = \begin{bmatrix} \color{red}{a_{11}} & \color{blue}{a_{12}} & \color{green}{a_{13}} \\ \color{red}{a_{21}} & \color{blue}{a_{22}} & \color{green}{a_{23}} \\ \color{red}{a_{31}} & \color{blue}{a_{32}} & \color{green}{a_{33}} \\ \color{red}{a_{41}} & \color{blue}{a_{42}} & \color{green}{a_{43}} \end{bmatrix}$$ where the entries are defined by the equations \begin{array}{rcrcrcrcrc} T(b_1) & = & \color{red}{a_{11}}\,e_1 &+&\color{red}{a_{21}}\,e_2 &+&\color{red}{a_{31}}\,e_3 &+&\color{red}{a_{41}}\,e_4 \\ T(b_2) & = & \color{blue}{a_{12}}\,e_1 &+&\color{blue}{a_{22}}\,e_2 &+&\color{blue}{a_{32}}\,e_3 &+&\color{blue}{a_{42}}\,e_4 \\ T(b_3) & = & \color{green}{a_{13}}\,e_1 &+&\color{green}{a_{23}}\,e_2 &+&\color{green}{a_{33}}\,e_3 &+&\color{green}{a_{43}}\,e_4 \end{array}

If I understand your question correctly, then the columns of this matrix is what you're asking for.

method $1$:

Find the change of basis matrix from $E$ to $B$:

$1+t=1(1)+1(t)+1(t^2)$

$t^2\;\;\;\;\,=0(1)+0(t)+1(t^2)$

$t\;\;\;\;\;\;= 0(1)+1(t)+0(t^2)$

tranpose the coefficients and put as columns in a $3x3$ matrix, you get: $$C = \left( {\begin{array}{*{20}{c}}1&0&0\\1&0&1\\0&1&0\end{array}} \right)$$

take $C$ and multiply it by a coordinate vector corosponding to each coefficient in the polynomial. for example:

$$a+bt+ct^2$$ has the coordinate vector $$\left( \begin{array}{l}a\\b\\c\end{array} \right)$$ which is :

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

so multiply $C$ times each vector and you will get:

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

now as was said above me, find general solution and you will get:

$$\,x=a\\y=c\\\;\;\;\;\,\,z=b-a$$

so any coordinate vector with respect to the basis $B$ his representation as a vercor in $R^3$ is :

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

example :

$$\left( \begin{array}{l}2\\1\\3\end{array} \right) = 2\left( \begin{array}{l}1\\1\\0\end{array} \right) + 3\left( \begin{array}{l}0\\0\\1\end{array} \right) + (1 - 2)\left( \begin{array}{l}0\\1\\0\end{array} \right)$$

Method 2

find the coefficient as mentioned above. and noticed that according to the basis $B = \{1+t,t^2,t\}$

the cooridnate vectors representantiom as a basis of $R^3$ is : $$\left\{ {\left( \begin{array}{l}1\\0\\0\end{array} \right) + \left( \begin{array}{l}0\\1\\0\end{array} \right),\left( \begin{array}{l}0\\0\\1\end{array} \right),\left( \begin{array}{l}0\\1\\0\end{array} \right)} \right\} = \left\{ {\left( \begin{array}{l}1\\1\\0\end{array} \right),\left( \begin{array}{l}0\\0\\1\end{array} \right),\left( \begin{array}{l}0\\1\\0\end{array} \right)} \right\}$$