# Findthe Matrix T of the following linear transformationT:R2(x−>R2(x)definedbyT(ax^2+bx+c)=2ax+b

Findthe Matrix T of the following linear transformation
$T\phantom{\rule{mediummathspace}{0ex}}:\phantom{\rule{mediummathspace}{0ex}}R2\left(x\right)\phantom{\rule{mediummathspace}{0ex}}->R2\left(x\right)\phantom{\rule{mediummathspace}{0ex}}defined\phantom{\rule{mediummathspace}{0ex}}by\phantom{\rule{mediummathspace}{0ex}}T\left(a{x}^{2}+bx+c\right)=2ax+b$
You can still ask an expert for help

## Want to know more about Matrix transformations?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Harper Brewer
Polynomials are functions that are defined by the numbers that multiply $1,x,{x}^{2},{x}^{3},....$ in the plots of the polynomial. That is, a polynomial of degree 2 or less, $a{x}^{2}+bx+c$, is defined by the numbers (or "coefficients") $\left(a,b,c\right)$.
If the base of ${R}_{2}\left[x\right]$ (vector space of polynomials of degree not greater than 2) is $\left({x}^{2},x,1\right)$, the matrix of the linear application will have in the first, second and third columns, respectively, the coefficients of $T\left({x}^{2}\right)$, $T\left(x\right)$ and $T\left(1\right)$, that is, the coefficients of
$2x=0\ast {x}^{2}+2\ast x+0\ast 1$ (first column is ,
$1=0\ast {x}^{2}+0\ast x+1\ast 1$ (second column is and
$0=0\ast {x}^{2}+0\ast x+0\ast 1$ (third column is .
as auxiliary calculus we have,
canonical basis $\left({x}^{2},x,1\right)$
$T\left({x}^{2}\right)=T\left(1{x}^{2}+0x+0\right)=2\ast 1\ast x+0=2x$
$T\left(x\right)=T\left(0{x}^{2}+x+0\right)=2\ast 0x+1=1$
$T\left(1\right)=T\left(0{x}^{2}+0x+1\right)=0$
solution $\left(2x,1,0\right)$