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 \({\mathbf{\text{p}}}_{{1}}{\quad\text{and}\quad}{\mathbf{\text{p}}}_{{2}}\in{\mathbb{\text{P}}}_{{2}}\) that spans the kernel of T and describes the range of sf "T".

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Expert Answer

Sadie Eaton
Answered 2020-11-21 Author has 28114 answers

The linear transformation T is defined as sf \({T}:{\mathbb{\text{P}}}_{{2}}\to{\mathbb{\text{R}}}^{2}{b}{y}{\left({\mathbf{\text{p}}}\right)}={\left[\begin{matrix}{\mathbf{\text{p}}}&{\left({0}\right)}\\{\mathbf{\text{p}}}&{\left({0}\right)}\end{matrix}\right]}\)
The set of polynomial p in \({\mathbb{\text{P}}}^{2}\ \text{such that}\ {T}{\left({\mathbf{\text{p}}}\right)}{0},\) 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 \({\mathbf{\text{p}}}_{{1}}{\left({t}\right)}={t}{\quad\text{and}\quad}{\mathbf{\text{p}}}_{{2}}{\left({t}\right)}=\) 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 \({\mathbf{\text{p}}}_{{1}}{\left({t}\right)}={t}{\quad\text{and}\quad}{\mathbf{\text{p}}}_{{2}}{\left({t}\right)}={t}^{2}\) spans the kernel of T.
The set of all polynomial contained in range of T is \({\left\lbrace{p}\in{V}:{T}{\left({\left({p}\right)}\right\rbrace}\right.}\)
The value of p (0) is always constant.
Thus, the range of T is given by \({\left\lbrace{\left[\begin{matrix}{c}\\{c}\end{matrix}\right]}:{c}{i}{s}\in{\mathbb{\text{R}}}\right\rbrace}.\)
The range of T is \({\left\lbrace{\left[\begin{matrix}{c}\\{c}\end{matrix}\right]}:{c}{i}{s}\in{\mathbb{\text{R}}}\right\rbrace}.\)

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