Linear Algebra and Its Applications, Exercise 2.4.13

Exercise 2.4.13. What is the rank of each of the following matrices:

A = \begin{bmatrix} 1&0&0&3 \\ 0&0&0&0 \\ 2&0&0&6 \end{bmatrix} \qquad A = \begin{bmatrix} 2&-2 \\ 2&-2 \end{bmatrix}

Express each matrix as a product of a column vector and row vector, A = uv^T.

Answer: We do Gaussian elimination on the first matrix by subtracting two times the first row from the third:

\begin{bmatrix} 1&0&0&3 \\ 0&0&0&0 \\ 2&0&0&6 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0&0&3 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix}

The resulting echelon matrix has one pivot, and thus the rank of A is 1. The matrix A can be expressed as the product of a column vector and row vector as follows:

A = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}\begin{bmatrix} 1&0&0&3 \end{bmatrix}

We do Gaussian elimination on the second matrix by subtracting the first row from the second:

\begin{bmatrix} 2&-2 \\ 2&-2 \end{bmatrix} \Rightarrow \begin{bmatrix} 2&-2 \\ 0&0 \end{bmatrix}

The resulting echelon matrix has one pivot, and thus the rank of A is 1. The matrix A can be expressed as the product of a column vector and row vector as follows:

A = \begin{bmatrix} 2 \\ 2 \end{bmatrix} \begin{bmatrix} 1&-1 \end{bmatrix}

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.

 Buy me a snack to sponsor more posts like this!

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

Leave a Reply

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

WordPress.com Logo

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

Twitter picture

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

Facebook photo

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

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s