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Chapter 4: Q12SE (page 191)

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

  1. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\). (Hint: Explain why every vector in the column space of \(AB\) is in the column space of \(A\).
  2. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} B\). (Hint: Use part (a) to study rank\({\left( {AB} \right)^T}\).)

Short Answer

Expert verified
  1. It is proved that \({\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} A\).
  2. It is proved that \({\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} B\).

Step by step solution

01

Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\)

a)

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

Consider \({\mathop{\rm y}\nolimits} \) is in \({\mathop{\rm Col}\nolimits} AB\), then \({\mathop{\rm y}\nolimits} = AB\) for some \({\mathop{\rm x}\nolimits} \). However, as \(AB{\mathop{\rm x}\nolimits} = A\left( {B{\mathop{\rm x}\nolimits} } \right)\), \({\mathop{\rm y}\nolimits} = A\left( {B{\mathop{\rm x}\nolimits} } \right)\), and \({\mathop{\rm y}\nolimits} \) is in \({\mathop{\rm Col}\nolimits} A\). Therefore, \({\mathop{\rm Col}\nolimits} AB\) is a subspace of \({\mathop{\rm Col}\nolimits} A\), and \({\mathop{\rm rank}\nolimits} AB = \dim {\mathop{\rm Col}\nolimits} AB \le \dim {\mathop{\rm Col}\nolimits} A = {\mathop{\rm rank}\nolimits} A\) (according to theorem 11).

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

02

Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} B\)

b)

The rank theoremstates that the dimensions of the column space and the row space of a \(m \times n\) matrix \(A\) are equal. This common dimension (the rank of \(A\)) also equals the number of pivot positions in \(A\) and satisfies the equation \({\mathop{\rm rank}\nolimits} A + \dim {\mathop{\rm Nul}\nolimits} A = n\).

\(\begin{array}{c}{\mathop{\rm rank}\nolimits} AB = {\mathop{\rm rank}\nolimits} {\left( {AB} \right)^T}\\ = {\mathop{\rm rank}\nolimits} {B^T}{A^T}\\ \le {\mathop{\rm rank}\nolimits} {B^T}\\ = {\mathop{\rm rank}\nolimits} B\end{array}\)

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

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Most popular questions from this chapter

Question 11: Let\(S\)be a finite minimal spanning set of a vector space\(V\). That is,\(S\)has the property that if a vector is removed from\(S\), then the new set will no longer span\(V\). Prove that\(S\)must be a basis for\(V\).

Let be a basis of\({\mathbb{R}^n}\). .Produce a description of an \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\)matrix A that implements the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\). Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)). (See Exercise 21.)

Suppose the solutions of a homogeneous system of five linear equations in six unknowns are all multiples of one nonzero solution. Will the system necessarily have a solution for every possible choice of constants on the right sides of the equations? Explain.

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 \(W\).

Consider the following two systems of equations:

\(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 1\\4{x_1} + {x_2} - 6{x_3} = 9\end{array}\) \(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 5\\4{x_1} + {x_2} - 6{x_3} = 45\end{array}\)

It can be shown that the first system of a solution. Use this fact and the theory from this section to explain why the second system must also have a solution. (Make no row operations.)
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