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Expert-verified Found in: Page 191 ### Linear Algebra and its Applications

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

See the step by step solution

## Step 1: 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 $${\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B \le n$$. ### Want to see more solutions like these? 