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Define by \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 1 \right)}\end{array}} \right)\). For instance, if \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + 5t + 7{t^2}\), then \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}3\\{15}\end{array}} \right)\).

  1. Show that \(T\) is a linear transformation. (Hint: For arbitrary polynomials p, q in \({{\mathop{\rm P}\nolimits} _2}\), compute \(T\left( {{\mathop{\rm p}\nolimits} + {\mathop{\rm q}\nolimits} } \right)\) and \(T\left( {c{\mathop{\rm p}\nolimits} } \right)\).)
  2. Find a polynomial p in \({{\mathop{\rm P}\nolimits} _2}\) that spans the kernel of \(T\), and describe the range of \(T\).

Short Answer

Expert verified
  1. It is proved that \(T\) is a linear transformation.
  2. The range of \(T\) is all of \({\mathbb{R}^2}\).

Step by step solution

01

State the condition for linear transformation

The conditions forlinear transformation\(T\)are as follows:

1.\(T\left( {{\mathop{\rm u}\nolimits} + {\mathop{\rm v}\nolimits} } \right) = T\left( {\mathop{\rm u}\nolimits} \right) + T\left( {\mathop{\rm v}\nolimits} \right)\) for all \({\mathop{\rm u}\nolimits} ,{\mathop{\rm v}\nolimits} \,\,{\mathop{\rm in}\nolimits} \,\,V\), and

2 \(T\left( {c{\mathop{\rm u}\nolimits} } \right) = cT\left( {\mathop{\rm u}\nolimits} \right)\) for all \({\mathop{\rm u}\nolimits} \,\,\,{\mathop{\rm in}\nolimits} \,\,V\) and all scalar \(c\).

02

Show that \(T\) is a linear transformation

a)

Suppose \({\mathop{\rm p}\nolimits} \) and \(q\) are arbitrary polynomials in \({{\mathop{\rm P}\nolimits} _2}\). Then

\(\begin{array}{c}T\left( {{\mathop{\rm p}\nolimits} + q} \right) = \left( {\begin{array}{*{20}{c}}{\left( {{\mathop{\rm p}\nolimits} + q} \right)\left( 0 \right)}\\{\left( {{\mathop{\rm p}\nolimits} + q} \right)\left( 1 \right)}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right) + {\mathop{\rm q}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 1 \right) + {\mathop{\rm q}\nolimits} \left( 1 \right)}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 1 \right)}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{{\mathop{\rm q}\nolimits} \left( 0 \right)}\\{{\mathop{\rm q}\nolimits} \left( 1 \right)}\end{array}} \right)\\ = T\left( {\mathop{\rm p}\nolimits} \right) + T\left( {\mathop{\rm q}\nolimits} \right)\end{array}\)

Consider \(c\) is any scalar. Then,

\(\begin{array}{c}T\left( {c{\mathop{\rm p}\nolimits} } \right) = \left( {\begin{array}{*{20}{c}}{\left( {c{\mathop{\rm p}\nolimits} } \right)\left( 0 \right)}\\{\left( {c{\mathop{\rm p}\nolimits} } \right)\left( 1 \right)}\end{array}} \right)\\ = c\left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 1 \right)}\end{array}} \right)\\ = cT\left( {\mathop{\rm p}\nolimits} \right)\end{array}\)

Therefore, \(T\) is a li\(T:{{\mathop{\rm P}\nolimits} _2} \to {\mathbb{R}^2}\)near transformation.

Thus, it is proved that \(T\) is a linear transformation.

03

 Determine polynomial \({\mathop{\rm p}\nolimits} \) in \({{\mathop{\rm P}\nolimits} _2}\) that spans the kernel of \(T\) 

b)

Any quadratic polynomial \({\mathop{\rm q}\nolimits} \) that has \({\mathop{\rm q}\nolimits} \left( 0 \right) = 0\) and \({\mathop{\rm q}\nolimits} \left( 1 \right) = 0\) will be included in the kernel of \(T\). Polynomial \({\mathop{\rm q}\nolimits} \) is then a multiple of \({\mathop{\rm p}\nolimits} \left( t \right) = t\left( {t - 1} \right)\).

04

Describe the range of T

It is given that any vector \(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\) is in \({\mathbb{R}^2}\).

The polynomial \({\mathop{\rm p}\nolimits} = {{\mathop{\rm x}\nolimits} _1} + \left( {{{\mathop{\rm x}\nolimits} _2} - {{\mathop{\rm x}\nolimits} _1}} \right)t\) have values such that \({\mathop{\rm p}\nolimits} \left( 0 \right) = {{\mathop{\rm x}\nolimits} _1}\) and \({\mathop{\rm p}\nolimits} \left( 1 \right) = {{\mathop{\rm x}\nolimits} _2}\).

Therefore, the range of \(T\) is all of \({\mathbb{R}^2}\).

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Most popular questions from this chapter

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

Let V be a vector space that contains a linearly independent set \(\left\{ {{u_{\bf{1}}},{u_{\bf{2}}},{u_{\bf{3}}},{u_{\bf{4}}}} \right\}\). Describe how to construct a set of vectors \(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\) in V such that \(\left\{ {{v_{\bf{1}}},{v_{\bf{3}}}} \right\}\) is a basis for Span\(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\).

Verify that rank \({{\mathop{\rm uv}\nolimits} ^T} \le 1\) if \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 3}\\5\end{array}} \right]\) and \({\mathop{\rm v}\nolimits} = \left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right]\).

Consider the polynomials , and \({p_{\bf{3}}}\left( t \right) = {\bf{2}}\) \({p_{\bf{1}}}\left( t \right) = {\bf{1}} + t,{p_{\bf{2}}}\left( t \right) = {\bf{1}} - t\)(for all t). By inspection, write a linear dependence relation among \({p_{\bf{1}}},{p_{\bf{2}}},\) and \({p_{\bf{3}}}\). Then find a basis for Span\(\left\{ {{p_{\bf{1}}},{p_{\bf{2}}},{p_{\bf{3}}}} \right\}\).

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show the coordinate mapping is one to one. (Hint: Suppose \({\left( {\bf{u}} \right)_B} = {\left( {\bf{w}} \right)_B}\) for some u and w in V, and show that \({\bf{u}} = {\bf{w}}\)).

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