 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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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 $${\left\lbrace{\left[\begin{matrix}{c}\\{c}\end{matrix}\right]}:{c}{i}{s}\in{\mathbb{\text{R}}}\right\rbrace}.$$