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Chapter 4: Q14E (page 191)

Question: In Exercises 13 and 14, assume that A is row equivalent to B. Find bases for NulA and Col A.

14. \(A = \left[ {\begin{array}{*{20}{c}}1&2&{ - 5}&{11}&{ - 3}\\2&4&{ - 5}&{15}&2\\1&2&0&4&5\\3&6&{ - 5}&{19}&{ - 2}\end{array}} \right]\), \(B = \left[ {\begin{array}{*{20}{c}}1&2&0&4&5\\0&0&5&{ - 7}&8\\0&0&0&0&{ - 9}\\0&0&0&0&0\end{array}} \right]\)

Short Answer

Expert verified

The basis for Col Ais \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\2\\1\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - 5}\\0\\{ - 5}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 3}\\2\\5\\{ - 2}\end{array}} \right]} \right\}\). The basis for NulAis \(\left\{ {\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\0\\{\frac{7}{5}}\\1\\0\end{array}} \right]} \right\}\).

Step by step solution

01

State the basis for NulA and Col A

The set of alllinear combinations of the columns of matrix A is Col A.It is called thecolumn space of A. Pivot columns are thebasis for Col A.

The set of all homogeneous equationsolutions,\(A{\bf{x}} = 0\), is NulA.It is called the null spaceof A.

02

Write the basis for Col A

To identify the pivot and the pivot position, observe the leftmost column (nonzero column) of the matrix, that is, the pivot column. At the top of this column, 1 is the pivot.

B = \(\left[ {\begin{array}{*{20}{c}} {\boxed1}&2&0&4&5 \\ 0&0&{\boxed5}&{ - 7}&8 \\ 0&0&0&0&{\boxed{ - 9}} \\ 0&0&0&0&0 \end{array}} \right]\)

The first, third and fifth columns have pivot elements.

The corresponding columns of matrix A are shown below:

\(\left[ {\begin{array}{*{20}{c}}1\\2\\1\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - 5}\\0\\{ - 5}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 3}\\2\\5\\{ - 2}\end{array}} \right]\)

The column space is shown below:

\({\rm{Col }}A = \left\{ {\left[ {\begin{array}{*{20}{c}}1\\2\\1\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - 5}\\0\\{ - 5}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 3}\\2\\5\\{ - 2}\end{array}} \right]} \right\}\)

Thus, the basis for Col Ais \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\2\\1\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - 5}\\0\\{ - 5}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 3}\\2\\5\\{ - 2}\end{array}} \right]} \right\}\).

03

Write the basis for Nul A

It is given that there are five columns in the given matrix.It means there should be five entries in vector x.

Thus, the equation \(A{\bf{x}} = 0\) can be written as shown below:

\(\begin{array}{c}Ax = 0\\\left[ {\begin{array}{*{20}{c}}1&2&0&4&5\\0&0&5&{ - 7}&8\\0&0&0&0&{ - 9}\\0&0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\end{array}\)

The system of equations obtained is shown below:

\(\begin{array}{c}{x_1} + 2{x_2} + 4{x_4} + 5{x_5} = 0\\5{x_3} - 7{x_4} + 8{x_5} = 0\\ - 9{x_5} = 0\end{array}\)

Or,

\({x_5} = 0\)

From the above equations, \({x_1}\),\({x_3}\), and \({x_5}\) are the pivot positions. So, \({x_1}\), \({x_3}\), and \({x_5}\)are the basic variables, and \({x_2}\)and\({x_4}\) are the free variables.

Let \({x_2} = a\), \({x_4} = b\).

Substitute the values\({x_2} = a\) and \({x_4} = b\) in the equation \({x_1} + 2{x_2} + 4{x_4} + 5{x_5} = 0\) to obtain the general solution.

\(\begin{array}{c}{x_1} + 2\left( a \right) + 4\left( b \right) + 5\left( 0 \right) = 0\\{x_1} = - 2a - 4b\end{array}\)

Substitute the values\({x_2} = a\) and \({x_4} = b\) in the equation \(5{x_3} - 7{x_4} + 8{x_5} = 0\) to obtain the general solution.

\(\begin{aligned}{}5{x_3} - 7\left( b \right) + 8\left( 0 \right) &= 0\\{x_3} &= \frac{7}{5}b\end{aligned}\)

Obtain the vector in the parametric form using \({x_1} = - 2a - 4b\), \({x_2} = a\), \[{x_3} = 7b\],\({x_4} = b\), and \({x_5} = 0\).

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2a - 4b}\\a\\{\frac{7}{5}b}\\b\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 2a}\\a\\0\\0\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 4b}\\0\\{\frac{7}{5}b}\\b\\0\end{array}} \right]\\ = a\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\\0\\0\end{array}} \right] + b\left[ {\begin{array}{*{20}{c}}{ - 4}\\0\\{\frac{7}{5}}\\1\\0\end{array}} \right]\\ = {x_2}\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\\0\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}{ - 4}\\0\\{\frac{7}{5}}\\1\\0\end{array}} \right]\end{array}\)

The basis of Nul A is shown below:

\({\rm{Nul }}A = \left\{ {\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\0\\{\frac{7}{5}}\\1\\0\end{array}} \right]} \right\}\)

Thus, the basis for NulAis \(\left\{ {\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\0\\{\frac{7}{5}}\\1\\0\end{array}} \right]} \right\}\).

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Most popular questions from this chapter

(calculus required) Define \(T:C\left( {0,1} \right) \to C\left( {0,1} \right)\) as follows: For f in \(C\left( {0,1} \right)\), let \(T\left( t \right)\) be the antiderivative \({\mathop{\rm F}\nolimits} \) of \({\mathop{\rm f}\nolimits} \) such that \({\mathop{\rm F}\nolimits} \left( 0 \right) = 0\). Show that \(T\) is a linear transformation, and describe the kernel of \(T\). (See the notation in Exercise 20 of Section 4.1.)

Consider the polynomials \({{\bf{p}}_{\bf{1}}}\left( t \right) = {\bf{1}} + t\), \({{\bf{p}}_{\bf{2}}}\left( t \right) = {\bf{1}} - t\), \({{\bf{p}}_{\bf{3}}}\left( t \right) = {\bf{4}}\), \({{\bf{p}}_{\bf{4}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}}\), and \({{\bf{p}}_{\bf{5}}}\left( t \right) = {\bf{1}} + {\bf{2}}t + {t^{\bf{2}}}\), and let H be the subspace of \({P_{\bf{5}}}\) spanned by the set \(S = \left\{ {{{\bf{p}}_{\bf{1}}},\,{{\bf{p}}_{\bf{2}}},\;{{\bf{p}}_{\bf{3}}},\,{{\bf{p}}_{\bf{4}}},\,{{\bf{p}}_{\bf{5}}}} \right\}\). Use the method described in the proof of the Spanning Set Theorem (Section 4.3) to produce a basis for H. (Explain how to select appropriate members of S.)

Let \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\). Find \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\) such that \(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\2&{ - 6}&8\end{array}} \right] = {{\mathop{\rm uv}\nolimits} ^T}\) .

In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work

\({\left( {{\bf{1}} - t} \right)^{\bf{2}}}\),\(t - {\bf{2}}{t^{\bf{2}}} + {t^{\bf{3}}}\),\({\left( {{\bf{1}} - t} \right)^{\bf{3}}}\)

Question 18: Suppose A is a \(4 \times 4\) matrix and B is a \(4 \times 2\) matrix, and let \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) represent a sequence of input vectors in \({\mathbb{R}^2}\).

  1. Set \({{\mathop{\rm x}\nolimits} _0} = 0\), compute \({{\mathop{\rm x}\nolimits} _1},...,{{\mathop{\rm x}\nolimits} _4}\) from equation (1), and write a formula for \({{\mathop{\rm x}\nolimits} _4}\) involving the controllability matrix \(M\) appearing in equation (2). (Note: The matrix \(M\) is constructed as a partitioned matrix. Its overall size here is \(4 \times 8\).)
  2. Suppose \(\left( {A,B} \right)\) is controllable and v is any vector in \({\mathbb{R}^4}\). Explain why there exists a control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) in \({\mathbb{R}^2}\) such that \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).
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