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Chapter 4: Q31E (page 191)

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

Justify the following equalities:

a.\({\rm{dim Row }}A{\rm{ + dim Nul }}A = n{\rm{ }}\)

b.\({\rm{dim Col }}A{\rm{ + dim Nul }}{A^T} = m\)

Consider the polynomials \({p_{\bf{1}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}},{p_{\bf{2}}}\left( t \right) = {\bf{1}} - {t^{\bf{2}}}\). Is \(\left\{ {{p_{\bf{1}}},{p_{\bf{2}}}} \right\}\) a linearly independent set in \({{\bf{P}}_{\bf{3}}}\)? Why or why not?

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

17. A submatrix of a matrix A is any matrix that results from deleting some (or no) rows and/or columns of A. It can be shown that A has rank \(r\) if and only if A contains an invertible \(r \times r\) submatrix and no longer square submatrix is invertible. Demonstrate part of this statement by explaining (a) why an \(m \times n\) matrix A of rank \(r\) has an \(m \times r\) submatrix \({A_1}\) of rank \(r\), and (b) why \({A_1}\) has an invertible \(r \times r\) submatrix \({A_2}\).

The concept of rank plays an important role in the design of engineering control systems, such as the space shuttle system mentioned in this chapter鈥檚 introductory example. A state-space model of a control system includes a difference equation of the form

\({{\mathop{\rm x}\nolimits} _{k + 1}} = A{{\mathop{\rm x}\nolimits} _k} + B{{\mathop{\rm u}\nolimits} _k}\)for \(k = 0,1,....\) (1)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(\left\{ {{{\mathop{\rm x}\nolimits} _k}} \right\}\) is a sequence of 鈥渟tate vectors鈥� in \({\mathbb{R}^n}\) that describe the state of the system at discrete times, and \(\left\{ {{{\mathop{\rm u}\nolimits} _k}} \right\}\) is a control, or input, sequence. The pair \(\left( {A,B} \right)\) is said to be controllable if

\({\mathop{\rm rank}\nolimits} \left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}& \cdots &{{A^{n - 1}}B}\end{array}} \right) = n\) (2)

The matrix that appears in (2) is called the controllability matrix for the system. If \(\left( {A,B} \right)\) is controllable, then the system can be controlled, or driven from the state 0 to any specified state \({\mathop{\rm v}\nolimits} \) (in \({\mathbb{R}^n}\)) in at most \(n\) steps, simply by choosing an appropriate control sequence in \({\mathbb{R}^m}\). This fact is illustrated in Exercise 18 for \(n = 4\) and \(m = 2\). For a further discussion of controllability, see this text鈥檚 website (Case study for Chapter 4).

Suppose the solutions of a homogeneous system of five linear equations in six unknowns are all multiples of one nonzero solution. Will the system necessarily have a solution for every possible choice of constants on the right sides of the equations? Explain.

(M) Show that \(\left\{ {t,sin\,t,cos\,{\bf{2}}t,sin\,t\,cos\,t} \right\}\) is a linearly independent set of functions defined on \(\mathbb{R}\). Start by assuming that

\({c_{\bf{1}}} \cdot t + {c_{\bf{2}}} \cdot sin\,t + {c_{\bf{3}}} \cdot cos\,{\bf{2}}t + {c_{\bf{4}}} \cdot sin\,t\,cos\,t = {\bf{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{5}} \right)\)

Equation (5) must hold for all real t, so choose several specific values of t (say, \(t = {\bf{0}},\,.{\bf{1}},\,.{\bf{2}}\)) until you get a system of enough equations to determine that the \({c_j}\) must be zero.
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