 # Find polynomial p_1 and p_2 in P^2 that spans the kernel of T and describes the range of sf "T". fortdefruitI 2020-11-20 Answered

Find polynomial ${\mathbf{\text{p}}}_{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathbf{\text{p}}}_{2}\in {\mathbb{\text{P}}}_{2}$ that spans the kernel of T and describes the range of sf "T".

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 Sadie Eaton

The linear transformation T is defined as sf $T:{\mathbb{\text{P}}}_{2}\to {\mathbb{\text{R}}}^{2}by\left(\mathbf{\text{p}}\right)=\left[\begin{array}{cc}\mathbf{\text{p}}& \left(0\right)\\ \mathbf{\text{p}}& \left(0\right)\end{array}\right]$
The set of polynomial p in are the members of kernel of T.
Therefore, any quadratic polynomial p with properties $\mathbf{\text{p}}\left(0\right)=0$ will be in kernel of T.
Thus, the two polynomials ${\mathbf{\text{p}}}_{1}\left(t\right)=t\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathbf{\text{p}}}_{2}\left(t\right)=$ wil be in kernel of T.
The polynomial t and ${t}^{2}$ linearly independent.
Thus, the linear combination of t and ${t}^{2}$ will be in kernel of T.
Hence, the polynomials ${\mathbf{\text{p}}}_{1}\left(t\right)=t\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathbf{\text{p}}}_{2}\left(t\right)={t}^{2}$ spans the kernel of T.
The set of all polynomial contained in range of T is $\left\{p\in V:T\left(\left(p\right)\right\}$
The value of p (0) is always constant.
Thus, the range of T is given by $\left\{\left[\begin{array}{c}c\\ c\end{array}\right]:cis\in \mathbb{\text{R}}\right\}.$
The range of T is $\left\{\left[\begin{array}{c}c\\ c\end{array}\right]:cis\in \mathbb{\text{R}}\right\}.$