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Q15SE

Expert-verifiedFound in: Page 191

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**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.)**

It is proved 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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