Izabella Ponce

2022-06-26

If
$A=\left[\begin{array}{ccc}1& -1& 2\\ -2& 1& -1\\ 1& 2& 3\end{array}\right]$
is the matrix representation of a linear transformation
$T:{P}_{2}\left(x\right)\to {P}_{2}\left(x\right)$
with respect to the bases $\left\{1-x,x\left(1-x\right),x\left(1+x\right)\right\}$ and $\left\{1,1+x,1+{x}^{2}\right\}$ then find T. What is the procedure to solve it?

crociandomh

Expert

Recall the the matrix of a transformation has as its columns the images of the domain basis vectors expressed relative to a basis of the codomain. Assuming that the first of the given bases is for the domain, this means that
$T\left[1-x\right]=1\cdot \left(1\right)-2\cdot \left(1+x\right)+1\cdot \left(1+{x}^{2}\right)={x}^{2}-2x.$
The other two columns give you $T\left[x\left(1-x\right)\right]$ and $T\left[x\left(1+x\right)\right]$, respectively. Now, express the general second-degree polynomial $a+bx+c{x}^{2}$ as a linear combination of the domain basis polynomials, i.e., as $\alpha \left(1-x\right)+\beta x\left(1-x\right)+\gamma x\left(1+x\right)$ and use linearity of $T$:
$T\left[a+bx+cy\right]=\alpha T\left[1-x\right]+\beta T\left[x\left(1-x\right)\right]+\gamma T\left[x\left(1+x\right)\right].$$T\left[a+bx+cy\right]=\alpha T\left[1-x\right]+\beta T\left[x\left(1-x\right)\right]+\gamma T\left[x\left(1+x\right)\right].$

Eden Solomon

Expert

The columns of the matrix of a transformation are the images of the basis vectors of the domain expressed relative to the basis of the codomain, i.e., it specifies a linear combination of the codomain basis vectors. So, the first column of A tells us that $T\left[1-x\right]=1\cdot \left(1\right)-2\cdot \left(1+x\right)+1\cdot \left(1+{x}^{2}\right)={x}^{2}-2x$, and so on. From that, you should be able to work out what T does to the general polynomial $a+bx+c{x}^{2}$. Alternatively, you might convert A to the standard basis and read the solution from that.