Question

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

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

Step-by-step solution

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