Chapter 4: Q8E (page 191)
Question: In Exercises5-8, find the steady-state vector.
8. \(\left[ {\begin{array}{*{20}{c}}{.7}&{.2}&{.2}\\0&{.2}&{.4}\\{.3}&{.6}&{.4}\end{array}} \right]\)
The steady-state vector is \({\mathop{\rm q}\nolimits} = \left[ {\begin{array}{*{20}{c}}{.4}\\{.2}\\{.4}\end{array}} \right]\).
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Let \(B = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\end{array}} \right),\,\left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{9}}\end{array}} \right)\,} \right\}\). Since the coordinate mapping determined by B is a linear transformation from \({\mathbb{R}^{\bf{2}}}\) into \({\mathbb{R}^{\bf{2}}}\), this mapping must be implemented by some \({\bf{2}} \times {\bf{2}}\) matrix A. Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)).
Use coordinate vector to test whether the following sets of poynomial span \({{\bf{P}}_{\bf{2}}}\). Justify your conclusions.
a. \({\bf{1}} - {\bf{3}}t + {\bf{5}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{5}}t - {\bf{7}}{t^{\bf{2}}}\), \( - {\bf{4}} + {\bf{5}}t - {\bf{6}}{t^{\bf{2}}}\), \({\bf{1}} - {t^{\bf{2}}}\)
b. \({\bf{5}}t + {t^{\bf{2}}}\), \({\bf{1}} - {\bf{8}}t - {\bf{2}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{4}}t + {\bf{2}}{t^{\bf{2}}}\), \({\bf{2}} - {\bf{3}}t\)
Suppose a \({\bf{4}} \times {\bf{7}}\) matrix A has four pivot columns. Is \({\bf{Col}}\,A = {\mathbb{R}^{\bf{4}}}\)? Is \({\bf{Nul}}\,A = {\mathbb{R}^{\bf{3}}}\)? Explain your answers.
In Exercise 18, Ais an \(m \times n\) matrix. Mark each statement True or False. Justify each answer.
18. a. If B is any echelon form of A, then the pivot columns of B form a basis for the column space of A.
b. Row operations preserve the linear dependence relations among the rows of A.
c. The dimension of the null space of A is the number of columns of A that are not pivot columns.
d. The row space of \({A^T}\) is the same as the column space of A.
e. If A and B are row equivalent, then their row spaces are the same.
In Exercise 5, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).
5. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{1}}\end{array}} \right)\)
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