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Chapter 4: Q15SE (page 191)

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

15. Let\(A\)be an\(m \times n\)matrix, and let\(B\)be a\(n \times p\)matrix such that\(AB = 0\). Show that\({\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B \le n\). (Hint: One of the four subspaces\({\mathop{\rm Nul}\nolimits} A\),\({\mathop{\rm Col}\nolimits} A,\,{\mathop{\rm Nul}\nolimits} B\), and\({\mathop{\rm Col}\nolimits} B\)is contained in one of the other three subspaces.)

Short Answer

Expert verified

It is proved that \({\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B \le n\).

Step by step solution

01

Show that \({\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B \le n\)

Let \(H\) be a subspace of a finite-dimensional vector space \(V\). According to Theorem 11,any linearly independent set can be expanded, if necessary, to a basis for \(H\). Also, \(H\) is finite-dimensional and \(\dim H \le \dim V\).

The equation \(AB = O\) demonstrates that every column of \(B\) is in \({\mathop{\rm Nul}\nolimits} A\). \({\mathop{\rm Col}\nolimits} B\) is a subspace of \({\mathop{\rm Nul}\nolimits} A\) because \({\mathop{\rm Nul}\nolimits} A\) is a subspace of \({\mathbb{R}^n}\). Also, all linear combinations of the columns of \(B\) are in \({\mathop{\rm Nul}\nolimits} A\).

According to theorem 11, in section 4.5, rank\(B = \dim {\mathop{\rm Col}\nolimits} B \le \dim {\mathop{\rm Col}\nolimits} A\). Use this inequality and the rank theorem applied to \(A\) as shown below:

\(\begin{array}{c}n = {\mathop{\rm rank}\nolimits} A + \dim {\mathop{\rm Nul}\nolimits} A\\ \ge {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\end{array}\)

Thus, it is proved that \({\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B \le n\).

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Most popular questions from this chapter

In Exercise 1, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

1. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{6}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{array}} \right)\)

Let \(V\) and \(W\) be vector spaces, and let \(T:V \to W\) be a linear transformation. Given a subspace \(U\) of \(V\), let \(T\left( U \right)\) denote the set of all images of the form \(T\left( {\mathop{\rm x}\nolimits} \right)\), where x is in \(U\). Show that \(T\left( U \right)\) is a subspace of \(W\).

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 \({\left( {\bf{x}} \right)_B}\)).

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

Exercises 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, \({c_{\bf{1}}}\),鈥�.,\({c_p}\))

\({c_{\bf{1}}}{{\bf{u}}_{\bf{1}}} + ..... + {c_p}{{\bf{u}}_p} = {\bf{0}}\) 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
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