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Chapter 4: Q21E (page 191)

In Exercises 21 and 22, mark each statement True or False.\(H = Span\left\{ {{b_1},...,{b_p}} \right\}\)Justify

each answer.

21. a.A single vector by itself is linearly dependent.

b. If , then\(\left\{ {{b_1},...,{b_p}} \right\}\)is a basis for H.

c.The columns of an invertible\(n \times n\)matrix form a basis for\({\mathbb{R}^n}\).

d. A basis is a spanning set that is as large as possible.

e. In some cases, the linear dependence relations among the columns of a matrix can be affected by certain elementary row operations on the matrix.

Short Answer

Expert verified

a. The statement is false.

b. The statement is false.

c. The statement is true.

d. The statement is false.

e. The statement is false.

Step by step solution

01

Mark the first statement true or false

The vectors are said to be linearly dependent if the equation \({c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} + ... + {c_p}{{\bf{v}}_p} = 0\) has a non-trivial solution, where \({c_1},{c_2},...,{c_p}\) are scalars, and not all the weights are zero.

For a single vector, the equation can be written as \(c{\bf{v}} = 0\).

According to the given statement, the equation should be linearly dependent, then the equation \(c{\bf{v}} = 0\) should have a non-trivial solution. It means weight \(c\) is non-zero and vector v is also non-zero.

As \(c\) and v are non-zero, so the equation \(c{\bf{v}} = 0\) cannot be satisfied. It is true only for zero vector.

So, the statement 鈥淎 single vector by itself is linearly dependent.鈥� is not always true.

Thus, statement (a) is false.

02

(b) Step 2: Mark the second statement true or false

The given statement 鈥淚f\(H = {\rm{Span}}\left\{ {{{\bf{b}}_1},...,{{\bf{b}}_p}} \right\}\), then\(\left\{ {{{\bf{b}}_1},...,{{\bf{b}}_p}} \right\}\)is a basis for H.鈥� can鈥檛 be true the reason is as follows:

If\(H = {\rm{Span}}\left\{ {{{\bf{b}}_1},...,{{\bf{b}}_p}} \right\}\), then the set of vectors\(\left\{ {{{\bf{b}}_1},...,{{\bf{b}}_p}} \right\}\)can be linearly dependent or independent.But\(\left\{ {{{\bf{b}}_1},...,{{\bf{b}}_p}} \right\}\) is the basis for H then the vectors are linearly independent.

Thus, statement (b) is false.

03

(c) Step 3: Mark the third statement true or false

Recall the invertible matrix theorem, which states that if the inverse of the matrix exists or the matrix is invertible, then the columns are linearly independent, and they form a basis for\({\mathbb{R}^n}\).

Thus, statement (c) is true.

04

Mark the fourth statement true or false

The given statement 鈥淎 basis is a spanning set that is as large as possible.鈥� cannot always be true.

The number of spanning sets can be more than the number of basis sets in\({\mathbb{R}^n}\). Not every spanning set can be used as a basis.

Thus, statement (d) is false.

05

(e) Step 5: Mark the fifth statement true or false

Consider matrices A and B, where matrix B is the row-reduced form of matrix A.

Matrix A can be different from matrix B, but the solution set of both the equations\(A{\bf{x}} = 0\)and\(B{\bf{x}} = 0\)will be the same.

So,the linear dependence relations among the columns of a matrix cannot be affected by certain elementary row operations on the matrix.

Thus, statement (e) is false.

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Most popular questions from this chapter

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).

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

21. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 2}&{ - 4.2}&{ - 4.8}&{ - 3.6}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

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. \({\bf{5}}t + {t^{\bf{2}}}\), \({\bf{1}} - {\bf{8}}t - {\bf{2}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{4}}t + {\bf{2}}{t^{\bf{2}}}\), \({\bf{2}} - {\bf{3}}t\)

(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 \({c_j}'{\mathop{\rm s}\nolimits} \).)

What would you have to know about the solution set of a homogenous system of 18 linear equations 20 variables in order to understand that every associated nonhomogenous equation has a solution? Discuss.
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