Linear Algebra and Its Applications, Exercise 3.2.10

Exercise 3.2.10. Discuss whether the projection matrix P from exercise 3.2.9 is invertible or not. Explain your answer.

Answer:The projection matrix

P = aa^T/a^Ta = \frac{1}{a_1a_1 + a_2a_2 + \dots + a_na_n} \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix}\begin{bmatrix} a_1&a_2&\dots&a_n \end{bmatrix}

= \frac{1}{a_1a_1 + a_2a_2 + \dots + a_na_n} \begin{bmatrix} a_1 \cdot a_1&a_1 \cdot a_2&\cdots&a_1 \cdot a_n \\ a_2 \cdot a_1&a_2 \cdot a_2&\cdots&a_2 \cdot a_n \\ \vdots&\vdots&\ddots&\vdots \\ a_n \cdot a_1&a_n \cdot a_2&\cdots&a_n \cdot a_n \end{bmatrix}

Note that each column i of the resulting matrix is equal to the original vector a = \left(a_1, a_2, \dots, a_n \right) multiplied by a scalar factor equal to a_i/a^Ta. The columns are therefore linearly dependent with each of them expressible as the first column times the factor a_i/a_1, and thus the rank of the projection matrix P is r = 1. Since the rank r < n the matrix P is singular and not invertible.

Also note that this is consistent with the geometry of the situation: Since P projects vectors from a higher-dimensional vector space onto a line, many points in the space will end up being mapped into a single point on a line. There is thus no way to reverse the mapping and map a point on the line onto a single unique vector in the higher-dimensional space.

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 and tagged , . 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