# Find the matrix A for the linear transformation T relative to the bases B = {1,x,x^2} and B' = {1,x,x^2,x^3} such that T(vecx) = Avecx?

## Let $T : {P}^{2} \mapsto {P}^{3}$ given by $4 x \cdot \vec{p}$ be a linear transformation. (I'm guessing $\vec{p}$ is a vector of polynomial terms?) I couldn't follow my professor in class when he did linear transformations involving nonstandard bases, because the notation was confusing... So I'm really not sure how to do this.

Apr 27, 2016

$A = \left[\begin{matrix}0 & 0 & 0 \\ 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{matrix}\right]$

#### Explanation:

The idea behind the notation of using $\vec{p}$ as both a vector and a polynomial is that when your vector space is ${P}^{n}$, your vectors are polynomials. Just as in ${\mathbb{R}}^{2}$, for example, you can consider a vector $\left(a , b\right)$ as being the sum $a \left(1 , 0\right) + b \left(0 , 1\right)$, you can think of the analogs in ${P}^{1}$ as being $\left(a , b\right)$ and $a \cdot 1 + b x$, where both vectors are being written in terms of the standard bases $\left\{\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right\}$ or $\left\{1 , x\right\}$.

You can have a nonstandard basis in ${\mathbb{R}}^{2}$, such as $B ' = \left\{\begin{matrix}1 & 1 \\ - 1 & 1\end{matrix}\right\}$, as long as the vectors are linearly independent. If we write a vector in terms of that basis, such as ${\left(a , b\right)}_{B '}$, then we should think of it as being the linear combination $a \left(1 , 1\right) + b \left(- 1 , 1\right)$ which would have the representation $\left(a - b , a + b\right)$ in the standard basis.

Similarly, you can have nonstandard bases in ${P}^{n}$, or in any vector space for that matter. One simply must remember that a vector in a given basis is a linear combination of the elements of that basis.

While this problem uses the standard bases for ${P}^{2}$ and ${P}^{3}$, we will instead treat them as if they are arbitrary bases, using the relabelings $B = \left\{1 , x , {x}^{2}\right\} = \left\{\vec{{u}_{0}} , \vec{{u}_{1}} , \vec{{u}_{2}}\right\}$ and $B ' = \left\{1 , x , {x}^{2} , {x}^{3}\right\} = \left\{\vec{{v}_{0}} , \vec{{v}_{1}} , \vec{{v}_{2}} , \vec{{v}_{3}}\right\}$.

For the problem itself, when we wish to find the matrix representation of a given transformation, all we need to do is see how the transformation acts on each member of the original basis and put that in terms of the target basis. The resulting vectors will be the column vectors of the matrix.

First, we see how $T$ acts on the members of $B$, and put that in terms of $B '$:

$T \left(\vec{{u}_{0}}\right) = 4 x = 4 \vec{{v}_{1}} = {\left[\begin{matrix}0 \\ 4 \\ 0 \\ 0\end{matrix}\right]}_{B '}$
$T \left(\vec{{u}_{1}}\right) = 4 {x}^{2} = 4 \vec{{v}_{2}} = {\left[\begin{matrix}0 \\ 0 \\ 4 \\ 0\end{matrix}\right]}_{B '}$
$T \left(\vec{{u}_{2}}\right) = 4 {x}^{3} = 4 \vec{{v}_{3}} = {\left[\begin{matrix}0 \\ 0 \\ 0 \\ 4\end{matrix}\right]}_{B '}$

Then we use those as the column vectors for our transformation matrix:

$A = \left[\begin{matrix}0 & 0 & 0 \\ 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{matrix}\right]$

Let's check to see if it works.

Given a polynomial $\vec{p} = {\left[\begin{matrix}a \\ b \\ c\end{matrix}\right]}_{B} = a + b x + c {x}^{2} \in {P}^{2}$ we have

$T \left(\vec{p}\right) = 4 x \left(a + b x + c {x}^{2}\right)$

$= 4 a x + 4 b {x}^{2} + 4 c {x}^{3}$

$= 0 \vec{{v}_{0}} + 4 a \vec{{v}_{1}} + 4 b \vec{{v}_{2}} + 4 c \vec{{v}_{3}}$

$= {\left[\begin{matrix}0 \\ 4 a \\ 4 b \\ 4 c\end{matrix}\right]}_{B '}$

$= \left[\begin{matrix}0 & 0 & 0 \\ 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{matrix}\right] {\left[\begin{matrix}a \\ b \\ c\end{matrix}\right]}_{B}$

$= A \vec{p}$