# Let T be a linear transformation from P_{2} text{into} P_{2} such that {T}{left({1}right)}={x},{T}{left({x}right)}={1}+{x}{quadtext{and}quad}{T}{left({x}^{2}right)}={1}+{x}+{x}^{2}. Find {T}{left({2}-{6}{x}+{x}^{2}right)}

Let T be a linear transformation from such that
$T\left(1\right)=x,T\left(x\right)=1+x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}T\left({x}^{2}\right)=1+x+{x}^{2}.$
Find $T\left(2-6x+{x}^{2}\right)$
You can still ask an expert for help

• 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

lamanocornudaW
Approach:
Let V and W be vector spaces. The function,
$T:V\to W$
is a linear transformation of V into W when the two properties below are thue for all u and v in V and for any scalar c.
$T\left(u+v\right)=T\left(u\right)+T\left(v\right)$
$T\left(cu\right)=cT\left(u\right)$
Calculation:
The value to be find is $T\left(2-6x+{x}^{2}\right).$
$2-6x+{x}^{2}={c}_{1}\left(1\right)+{c}_{2}\left(x\right)+{c}_{3}{x}^{2}$
Compare the coefficient of same power.
${c}_{1}=2$

${c}_{3}=1$
Consider a linear transformation T such that,
$v={c}_{1}{v}_{1}+{c}_{2}{v}_{2}+\dots +{c}_{n}{v}_{n}$
Then,
$T\left(v\right)=T\left({c}_{1}{v}_{1}+{c}_{2}{v}_{2}+\dots +{c}_{n}{v}_{n}\right)$
$={c}_{1}T\left({v}_{1}\right)+{c}_{2}T\left({v}_{2}\right)+\dots +{c}_{n}T\left({v}_{n}\right)$
So,
$T\left(2-6x+{x}^{2}\right)=2T\left(1\right)-6T\left(x\right)+T\left({x}^{2}\right)\dots \left(1\right)$
Subtitute x for $T\left(1\right),\left(1+x\right)f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}T\left(x\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(1+x+{x}^{2}\right)f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}T\left({x}^{2}\right)$ in equation (1).
$T\left(2-6x+{x}^{2}=2\left(x\right)-6\left(x+1\right)+\left(1+x+{x}^{2}\right)$
$2x-6x-6+1+x+{x}^{2}$
$=-5-3x+{x}^{2}$
Therefore, the value of $T\left(2-6x+{x}^{2}\right)is-5-3x+{x}^{2}.$