Linear Algebra and Its Applications, Exercise 3.3.21

Exercise 3.3.21. Given three vectors a_1, a_2, and b, and the two lines L_1 through the origin and a_1 and L_2 through b in the direction of a_2, we want to find scalar values x_1 and x_2 such that the distance \| x_1a_1 - x_2a_2 - b \| between the points x_1a_1 and b + x_2a_2 is at a minimum. Write down equations for x_1 and x_2. Solve the equations for x = (x_1, x_2) when a_1 = (1, 1, 0), a_2 = (0, 1, 0), and b = (2, 1, 4).

Answer: We approach the problem by taking partial derivatives of \| x_1a_1 - x_2a_2 - b \|^2 with respect to x_1 and x_2 and then setting those partial derivatives to zero. We have

\| x_1a_1 - x_2a_2 - b \|^2 = (x_1a_1 - x_2a_2 - b)^T(x_1a_1 - x_2a_2 - b)

We take advantage of the facts that for two vectors v and w we have v^Tw = w^Tv (the inner product is the same no matter in which order we calculate it) and (v-w)^T = v^T - w^T (the transpose of a difference is the same as the difference of the transposes). We then have

\| x_1a_1 - x_2a_2 - b \|^2 = (x_1a_1^T - x_2a_2^T - b^T)(x_1a_1 - x_2a_2 - b)

= x_1^2a_1^Ta_1 - x_1x_2a_1^Ta_2 - x_1a_1^Tb - x_2x_1a_2^Ta_1 + x_2^2a_2^Ta_2

+ x_2a_2^Tb - x_1b^Ta_1 + x_2b^Ta_2 + b^Tb

= x_1^2a_1^Ta_1 + x_2^2a_2^Ta_2 - 2x_1x_2a_1^Ta_2 - 2x_1a_1^Tb + 2x_2a_2^Tb + b^Tb

Next we take the partial derivatives with respect to x_1:

\frac{\partial}{\partial x_1} \| x_1a_1 - x_2a_2 - b \|^2

= \frac{\partial}{\partial x_1} (x_1^2a_1^Ta_1 + x_2^2a_2^Ta_2 - 2x_1x_2a_1^Ta_2 - 2x_1a_1^Tb + 2x_2a_2^Tb + b^Tb)

= 2x_1a_1^Ta_1 - 2x_2a_1^Ta_2 - 2a_1^Tb

and with respect to x_2:

\frac{\partial}{\partial x_2} \| x_1a_1 - x_2a_2 - b \|^2

= \frac{\partial}{\partial x_2} (x_1^2a_1^Ta_1 + x_2^2a_2^Ta_2 - 2x_1x_2a_1^Ta_2 - 2x_1a_1^Tb + 2x_2a_2^Tb + b^Tb)

= 2x_2a_2^Ta_2 - 2x_1a_1^Ta_2 + 2a_2^Tb

Equating the partial derivatives to zero gives us the following system of equations (after simplifying by dividing by 2 and rearranging terms):

\begin{bmatrix} a_1^Ta_1&-a_1^Ta_2 \\ -a_1^Ta_2&a_2^Ta_2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} a_1^Tb \\ -a_2^Tb \end{bmatrix}

When a_1 = (1, 1, 0), a_2 = (0, 1, 0), and b = (2, 1, 4) we have a_1^Ta_1 = 2, a_2^Ta_2 = 1, a_1^Ta_2 = 1, a_1^Tb = 3, and a_2^Tb = 1. The system of equations is then

\begin{bmatrix} 2&-1 \\ -1&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}

We can solve this by multiplying both sides of the equation on the left by

\begin{bmatrix} 2&-1 \\ -1&1 \end{bmatrix}^{-1} = \frac{1}{2 - (-1)(-1)} \begin{bmatrix} 1&1 \\ 1&2 \end{bmatrix} = \begin{bmatrix} 1&1 \\ 1&2 \end{bmatrix}

to get

\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 1&1 \\ 1&2 \end{bmatrix} \begin{bmatrix} 3 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}

The solution is thus x = (2, 1).

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, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.

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