Linear Algebra and Its Applications, Exercise 2.3.8

Exercise 2.3.8. Describe the column space of the matrix

$A = \begin{bmatrix} 1&2 \\ 3&6 \end{bmatrix}$

and give a basis for it. Do the same for $A^2$ .

Answer:The second column of $A$ is twice the first column, so that the two vectors are linearly dependent. The column space consists of any vector of the form $c (1, 3)$ where $c$ is any real number; geometrically the column space is a line passing through the origin and the point $(1, 3)$ . The vector $(1, 3)$ serves as a basis for the space.

We have

$A^2 = \begin{bmatrix} 1&2 \\ 3&6 \end{bmatrix} \begin{bmatrix} 1&2 \\ 3&6 \end{bmatrix} = \begin{bmatrix} 7&14 \\ 21&42 \end{bmatrix}$

Again we have the second column equal to twice the first, so the two vectors are linearly dependent. Also, we have $(7, 21) = 7(1, 3)$ so that the first column of $A^2$ is a linear combination of the first column of $A$ . The matrix $A^2$ therefore has the same column space as $A$ and the vector $(1, 3)$ can serve as its basis.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

This entry was posted in linear algebra. Bookmark the permalink.

Leave a Reply Cancel reply

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Cancel

Connecting to %s