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Chapter 4: Q32E (page 191)

Exercise 32 reveal an important connection between linear independence and linear transformations and provide practice using the definition of linear dependence. Let V and W be vector spaces, let \(T:V \to W\) be a linear transformation, and let \(\left\{ {{v_{\bf{1}}},...,{v_p}} \right\}\) be a subset of V.

32. Suppose that T is a one-to-one transformation, so that an equation \(T\left( u \right) = T\left( v \right)\) always implies \(u = v\). Show that if the set of images \(\left\{ {T\left( {{v_{\bf{1}}}} \right),...,T\left( {{v_p}} \right)} \right\}\) is linearly dependent, then \(\left\{ {{v_{\bf{1}}},...,{v_p}} \right\}\) is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set (because in this case the set of images cannot be linearly dependent).

Short Answer

Expert verified

The set \(\left\{ {{v_1},...,{v_p}} \right\}\) is linearly dependent.

Step by step solution

01

Write the given statement

Suppose \(\left\{ {T\left( {{v_1}} \right),...,T\left( {{v_p}} \right)} \right\}\) is linearly dependent.

02

Use the definition of linear dependence

There exist scalars \({c_1},...,{c_p}\) that are not all zeros, such that

\({c_1}T\left( {{v_1}} \right) + ... + {c_p}T\left( {{v_p}} \right) = 0\).

Since T is a linear transformation,

\(\begin{array}{c}T\left( {{c_1}{v_1} + ... + {c_p}{v_p}} \right) = 0\\ = T\left( 0 \right).\end{array}\)

Since T is one-to-one, \({c_1}{v_1} + ... + {c_p}{v_p} = 0\).

Since not all scalars are zeros.

03

Draw a conclusion

Thus, \(\left\{ {{v_1},...,{v_p}} \right\}\) is linearly dependent.

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Most popular questions from this chapter

Suppose a nonhomogeneous system of nine linear equations in ten unknowns has a solution for all possible constants on the right sides of the equations. Is it possible to find two nonzero solutions of the associated homogeneous system that are not multiples of each other? Discuss.

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

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

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’s 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 “state vectors” 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’s website (Case study for Chapter 4).

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

13. Show that if \(P\) is an invertible \(m \times m\) matrix, then rank\(PA\)=rank\(A\).(Hint: Apply Exercise12 to \(PA\) and \({P^{ - 1}}\left( {PA} \right)\).)

Let \(A\) be an \(m \times n\) matrix of rank \(r > 0\) and let \(U\) be an echelon form of \(A\). Explain why there exists an invertible matrix \(E\) such that \(A = EU\), and use this factorization to write \(A\) as the sum of \(r\) rank 1 matrices. [Hint: See Theorem 10 in Section 2.4.]
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