Question

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

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

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 setcan 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 \({\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} A\).

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 \({\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} B\).