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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 $$m \times n$$. 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 $$A$$.Show that if $$B$$ is $$n \times p$$, then rank$$AB \le {\mathop{\rm rank}\nolimits} B$$. (Hint: Use part (a) to study rank$${\left( {AB} \right)^T}$$.)

1. It is proved that $${\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} A$$.
2. It is proved that $${\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} B$$.
See the step by step solution

## Step 1: Show that if $$B$$ is $$n \times p$$, then rank$$AB \le {\mathop{\rm rank}\nolimits} A$$

a)

According to the theorem 11 let $$H$$ 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 $${\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} A$$.

## Step 2: Show that if $$B$$ is $$n \times p$$, then rank$$AB \le {\mathop{\rm rank}\nolimits} B$$

b)

The rank theorem states 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 $${\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} B$$. ### Want to see more solutions like these? 