Chapter 4: Q16E (page 191)
If A is a \({\bf{6}} \times {\bf{4}}\) matrix, what is the smallest possible dimension of Null A?
Short Answer
The smallest possible dimension of Null A is 0.
Chapter 4: Q16E (page 191)
If A is a \({\bf{6}} \times {\bf{4}}\) matrix, what is the smallest possible dimension of Null A?
The smallest possible dimension of Null A is 0.
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Get started for freeExercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).
Show that a subset \(\left\{ {{{\bf{u}}_1},...,{{\bf{u}}_p}} \right\}\) in V is linearly independent if and only if the set of coordinate vectors \(\left\{ {{{\left( {{{\bf{u}}_{\bf{1}}}} \right)}_B},.....,{{\left( {{{\bf{u}}_p}} \right)}_B}} \right\}\) is linearly independent in \({\mathbb{R}^n}\)(Hint: Since the coordinate mapping is one-to-one, the following equations have the same solutions, \({c_{\bf{1}}}\),….,\({c_p}\))
\({c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p} = {\bf{0}}\) The zero vector V
\({\left( {{c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p}} \right)_B} = {\left( {\bf{0}} \right)_B}\) The zero vector in \({\mathbb{R}^n}\)a
Find a basis for the set of vectors in\({\mathbb{R}^{\bf{3}}}\)in the plane\(x + {\bf{2}}y + z = {\bf{0}}\). (Hint:Think of the equation as a “system” of homogeneous equations.)
If a \({\bf{3}} \times {\bf{8}}\) matrix A has a rank 3, find dim Nul A, dim Row A, and rank \({A^T}\).
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).
Which of the subspaces \({\rm{Row }}A\) , \({\rm{Col }}A\), \({\rm{Nul }}A\), \({\rm{Row}}\,{A^T}\) , \({\rm{Col}}\,{A^T}\) , and \({\rm{Nul}}\,{A^T}\) are in \({\mathbb{R}^m}\) and which are in \({\mathbb{R}^n}\) ? How many distinct subspaces are in this list?.
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