Answer to Question #108061 in Linear Algebra for Anuja

Question

let T:P2-->P2 defined by T(a+bx+cx^2)=(2b+3cx+(a-b)x^2) check T is linear trnsfrmtn.find matrix wrt bases B={x2,x2+x,x2+x+1} ,B2={1,x,x2}.find kernel of T

Answer 1

$T(a+bx+cx^2)=2b+3cx+(a-b)x^2$

It should satisfy

$(i)T(u+v)=T(u)+T(v)$

$(ii)T(mu)=mT(u)$

$u=f+ex+dx^2;v=i+hx+gx^2$

$(i)T(u+v)=T((i+f)+(e+h)x+(d+g)x^2)$

$=2(e+h)+3(d+g)x+(i+f-c-h)x^2$

$=[2e+3dx+(f-e)x^2]+[2h+3gx+(i-h)x^2]$

$=T(u)+T(v)$

$(ii)T(mu)=T(mf+mex+mdx^2)$

$=2me+3mdx+(mf-me)x^2$

$=m(2e+3dx+(f-e)x^2)$

$=mT(u)$

So,T is a linear Transformation.

$B_1=({x^2,x^2+x,x^2+x+1})$

$T(x^2)=3x=3(x^2+x)-3(x^2)+0.(x^2+x+1)$

$T(x^2+x)=2+3x-x^2=(-4).x^2+1.(x^2+x)+2.(x^2+x+1)$

$T(x^2+x+1)=2+3x=(-3).x^2+1.(x^2+x)+2.(x^2+x+1)$

The coefficient of T(x²) is $\begin{pmatrix} -3 \\ 3\\ 0 \end{pmatrix}$

The coefficient of T(x²+x) is $\begin{pmatrix} -4 \\ 1\\ 2 \end{pmatrix}$

The coefficient of T(x²+x+1) is $\begin{pmatrix} -3 \\ 1\\ 2 \end{pmatrix}$

So the matrix is $\begin{bmatrix} -3& -4 &-3\\ 3 & 1&1\\ 0&2&2 \end{bmatrix}$

$B_2=(1,x,x^2)$

$T(1)=x^2$

$T(x)=2-x^2$

$T(x^2)=3x$

The coefficient of T(1) is $\begin{pmatrix} 0 \\ 0\\ 1 \end{pmatrix}$

The coefficient of T(x) is $\begin{pmatrix} 2 \\ 0\\ -1 \end{pmatrix}$

The coefficient of T(x²) is $\begin{pmatrix} 0 \\ 3\\ 0 \end{pmatrix}$

So,the matrix is $\begin{bmatrix} 0 & 2 &0\\ 0 & 0&3\\ 1&-1&0 \end{bmatrix}$

Ker T={v: $T(v)=0$ }

$2b+3cx+(a-b)x^2=0$

$2b=0;3c=0;(a-b)=0$

$b=0,c=0,a=b=0$

$a=b=c=0$

So,v=0

Ker T={0}